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== 7.6#h(0.6em)Solve Rational Inequalities

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Find the value of #math.equation(block: false, alt: "x minus 5")[$x − 5$] when ⓐ #math.equation(block: false, alt: "x equals 6")[$x = 6$] ⓑ #math.equation(block: false, alt: "x equals −3")[$x = −3$] ⓒ #math.equation(block: false, alt: "x equals 5 .")[$x = 5 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
ⓐ 1; ⓑ −8; ⓒ 0
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "8 minus 2 x less than 12 .")[$8 − 2 x < 12 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x greater than −2")[$x > −2$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write in interval notation: #math.equation(block: false, alt: "−3 less than or equal to x less than 5 .")[$−3 ≤ x < 5 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "[ −3 , 5 close parenthesis")[$[ −3 , #h(0.2em) 5 )$]
]
]


=== Solve Rational Inequalities

We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same with one major exception. When we multiplied or divided by a negative number, the inequality sign reversed.

Having just learned to solve rational equations we are now ready to solve rational inequalities. A #strong[rational inequality] is an inequality that contains a rational expression.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Rational Inequality]

A #strong[rational inequality] is an inequality that contains a rational expression.
]

Inequalities such as #math.equation(block: false, alt: "the fraction 3 over 2 x greater than 1 , the fraction 2 x over x minus 3 less than 4 , the fraction 2 x minus 3 over x minus 6 greater than or equal to x ,")[$frac(3, 2 x) > 1 , #h(0.5em) frac(2 x, x − 3) < 4 , #h(0.5em) frac(2 x − 3, x − 6) ≥ x ,$] and #math.equation(block: false, alt: "the fraction 1 over 4 minus the fraction 2 over x squared less than or equal to the fraction 3 over x")[$frac(1, 4) − frac(2, x^(2)) ≤ frac(3, x)$] are rational inequalities as they each contain a rational expression.

When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

When we solve an equation and the result is #math.equation(block: false, alt: "x equals 3 ,")[$x = 3 ,$] we know there is one solution, which is 3.

When we solve an inequality and the result is #math.equation(block: false, alt: "x greater than 3 ,")[$x > 3 ,$] we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a #strong[zero partition number] and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

#figure(figph[This figure shows the solution, the interval 3 to infinity, of the inequality x is greater than 3 on a number line. The values range from negative 5 to 5 on the number line. The inequality is modeled by an open parenthesis at the zero partition number 3 and shading the right.], alt: "This figure shows the solution, the interval 3 to infinity, of the inequality x is greater than 3 on a number line. The values range from negative 5 to 5 on the number line. The inequality is modeled by an open parenthesis at the zero partition number 3 and shading the right.", caption: none)

To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right.

Next we determine the zero partition numbers to use to divide the number line into intervals. A #strong[zero partition number] is a number which make the rational expression zero or undefined.

We then will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

#examplebox("Example 1")[][
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction x minus 1 over x plus 3 greater than or equal to 0 .")[$frac(x − 1, x + 3) ≥ 0 .$]

#solutionbox[
#strong[Step 1.] Write the inequality as one quotient on the left and zero on the right.

Our inequality is in this form. #math.equation(block: false, alt: "the fraction x minus 1 over x plus 3 greater than or equal to 0")[$#h(5em) frac(x − 1, x + 3) ≥ 0$]

#strong[Step 2.] Determine the zero partition numbers—the points where the rational expression will be zero or undefined.

The rational expression will be zero when the numerator is zero. Since #math.equation(block: false, alt: "x minus 1 equals 0")[$x − 1 = 0$] when #math.equation(block: false, alt: "x equals 1 ,")[$x = 1 ,$] then #math.equation(block: false, alt: "1")[$1$] is a zero partition number.

The rational expression will be undefined when the denominator is zero. Since #math.equation(block: false, alt: "x plus 3 equals 0")[$x + 3 = 0$] when #math.equation(block: false, alt: "x equals −3 ,")[$x = −3 ,$] then #math.equation(block: false, alt: "−3")[$−3$] is a zero partition number.

The zero partition numbers are 1 and #math.equation(block: false, alt: "−3 .")[$−3 .$]

#strong[Step 3.] Use the zero partition numbers to divide the number line into intervals.

#figure(figph[This figure shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0.], alt: "This figure shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0.", caption: none)

The number line is divided into three intervals:

#math.equation(block: true, alt: "open parenthesis − ∞ , −3 close parenthesis open parenthesis −3 , 1 close parenthesis open parenthesis 1 , ∞ close parenthesis")[$#h(10.3em) ( "−" ∞ , −3 ) #h(5em) ( −3 , 1 ) #h(5.5em) ( 1 , ∞ )$]

#strong[Step 4.] Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

#math.equation(block: true, alt: "Interval open parenthesis − ∞ , −3 close parenthesis")[$"Interval" #h(0.2em) ( "−" ∞ , −3 )$]The number #math.equation(block: false, alt: "−4")[$−4$] is in the interval #math.equation(block: false, alt: "open parenthesis − ∞ , −3 close parenthesis .")[$( "−" ∞ , −3 ) .$] Test #math.equation(block: false, alt: "x equals −4")[$x = −4$] in the expression in the numerator and the denominator.

#figure(figph[This figure labels the expression, x minus 1, as the “numerator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 5. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 1. It labels the result “negative”.], alt: "This figure labels the expression, x minus 1, as the “numerator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 5. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 1. It labels the result “negative”.", caption: none)

Above the number line, mark the factor #math.equation(block: false, alt: "x minus 1")[$x − 1$] negative and mark the factor #math.equation(block: false, alt: "x plus 3")[$x + 3$] negative.

Since a negative divided by a negative is positive, mark the quotient positive in the interval #math.equation(block: false, alt: "open parenthesis − ∞ , −3 close parenthesis .")[$( "−" ∞ , −3 ) .$]

#figure(figph[This figure shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is negative, which is positive. It shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3.], alt: "This figure shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is negative, which is positive. It shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3.", caption: none)

#math.equation(block: true, alt: "Interval open parenthesis −3 , 1 close parenthesis")[$"Interval" #h(0.2em) ( −3 , 1 )$]The number 0 is in the interval #math.equation(block: false, alt: "open parenthesis −3 , 1 close parenthesis .")[$( −3 , 1 ) .$] Test #math.equation(block: false, alt: "x equals 0 .")[$x = 0 .$]

#figure(figph[This figure labels the expression, x minus 1, as the “numerator”. It shows that when 0 is substituted into the expression for x, the result is negative 1. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 0 is substituted into the expression for x, the result is 3. It labels the result “positive”.], alt: "This figure labels the expression, x minus 1, as the “numerator”. It shows that when 0 is substituted into the expression for x, the result is negative 1. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 0 is substituted into the expression for x, the result is 3. It labels the result “positive”.", caption: none)

Above the number line, mark the factor #math.equation(block: false, alt: "x minus 1")[$x − 1$] negative and mark #math.equation(block: false, alt: "x plus 3")[$x + 3$] positive.

Since a negative divided by a positive is negative, the quotient is marked negative in the interval #math.equation(block: false, alt: "open parenthesis −3 , 1 close parenthesis .")[$( −3 , 1 ) .$]

#figure(figph[This figure shows a shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1.], alt: "This figure shows a shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1.", caption: none)

#math.equation(block: true, alt: "Interval open parenthesis 1 , ∞ close parenthesis")[$"Interval" #h(0.2em) ( 1 , ∞ )$]The number 2 is in the interval #math.equation(block: false, alt: "open parenthesis 1 , ∞ close parenthesis .")[$( 1 , ∞ ) .$] Test #math.equation(block: false, alt: "x equals 2 .")[$x = 2 .$]

#figure(figph[This figure labels the expression, x minus 1, as the “numerator”. It shows that when 2 is substituted into the expression for x, the result is 1. It labels the result as “positive”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 2 is substituted into the expression for x, the result is 5. It labels the result “positive”.], alt: "This figure labels the expression, x minus 1, as the “numerator”. It shows that when 2 is substituted into the expression for x, the result is 1. It labels the result as “positive”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 2 is substituted into the expression for x, the result is 5. It labels the result “positive”.", caption: none)

Above the number line, mark the factor #math.equation(block: false, alt: "x minus 1")[$x − 1$] positive and mark #math.equation(block: false, alt: "x plus 3")[$x + 3$] positive.

Since a positive divided by a positive is positive, mark the quotient positive in the interval #math.equation(block: false, alt: "open parenthesis 1 , ∞ close parenthesis .")[$( 1 , ∞ ) .$]

#figure(figph[The figure shows that in the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line is divided into intervals by zero partition numbers at negative 3 and 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.], alt: "The figure shows that in the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line is divided into intervals by zero partition numbers at negative 3 and 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.", caption: none)

#strong[Step 5.] Determine the intervals where the inequality is correct. Write the solution in interval notation.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals #math.equation(block: false, alt: "open parenthesis − ∞ , −3 close parenthesis")[$( "−" ∞ , −3 )$] and #math.equation(block: false, alt: "open parenthesis 1 , ∞ close parenthesis")[$( 1 , ∞ )$] are solutions.

But what about the zero partition numbers?

The zero partition number #math.equation(block: false, alt: "x equals −3")[$x = −3$] makes the denominator 0, so it must be excluded from the solution and we mark it with a parenthesis.

The zero partition number #math.equation(block: false, alt: "x equals 1")[$x = 1$] makes the whole rational expression 0. The inequality requires that the rational expression be greater than or equal to 0. So, 1 is part of the solution and we will mark it with a bracket.

#figure(figph[The number line is divided into intervals by zero partition numbers at negative 3 and 1. A closed parenthesis is used at 3 and an open bracket is used at 1. The number is shaded to the left of 3 and to the right of 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.], alt: "The number line is divided into intervals by zero partition numbers at negative 3 and 1. A closed parenthesis is used at 3 and an open bracket is used at 1. The number is shaded to the left of 3 and to the right of 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.", caption: none)

Recall that when we have a solution made up of more than one interval we use the union symbol, #math.equation(block: false, alt: "union ,")[$∪ ,$] to connect the two intervals. The solution in interval notation is #math.equation(block: false, alt: "open parenthesis − ∞ , −3 close parenthesis union [ 1 , ∞ close parenthesis .")[$( "−" ∞ , −3 ) ∪ [ 1 , ∞ ) .$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction x minus 2 over x plus 4 greater than or equal to 0 .")[$frac(x − 2, x + 4) ≥ 0 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 close parenthesis union [ 2 , ∞ close parenthesis")[$( "−" ∞ , −4 ) ∪ [ 2 , ∞ )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction x plus 2 over x minus 4 greater than or equal to 0 .")[$frac(x + 2, x − 4) ≥ 0 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −2 ] union open parenthesis 4 , ∞ close parenthesis")[$( "−" ∞ , −2 ] ∪ ( 4 , ∞ )$]
]
]

We summarize the steps for easy reference.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a rational inequality.]

+ Write the inequality as one quotient on the left and zero on the right.
+ Determine the zero partition numbers–the points where the rational expression will be zero or undefined.
+ Use the zero partition numbers to divide the number line into intervals.
+ Test a value in each interval. Above the number line show the sign of each factor of the numerator and denominator in each interval. Below the number line show the sign of the quotient.
+ Determine the intervals where the inequality is correct. Write the solution in interval notation.
]

The next example requires that we first get the rational inequality into the correct form.

#examplebox("Example 2")[][
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 4 x over x minus 6 less than 1 .")[$frac(4 x, x − 6) < 1 .$]

#solutionbox[
#figure(table(
  columns: 3, align: left, inset: 6pt,
  [], [#math.equation(block: false, alt: "the fraction 4 x over x minus 6 less than 1")[$frac(4 x, x − 6) < 1$]], [],
  [Subtract 1 to get zero on the right.], [#math.equation(block: false, alt: "the fraction 4 x over x minus 6 minus 1 less than 0")[$frac(4 x, x − 6) − 1 < 0$]], [],
  [Rewrite 1 as a fraction using the LCD.], [#math.equation(block: false, alt: "the fraction 4 x over x minus 6 minus the fraction x minus 6 over x minus 6 less than 0")[$frac(4 x, x − 6) − frac(x − 6, x − 6) < 0$]], [],
  [Subtract the numerators and place the #linebreak() difference over the common denominator.], [#math.equation(block: false, alt: "the fraction 4 x minus open parenthesis x minus 6 close parenthesis over x minus 6 less than 0")[$frac(4 x − ( x − 6 ), x − 6) < 0$]], [],
  [Simplify.], [#math.equation(block: false, alt: "the fraction 3 x plus 6 over x minus 6 less than 0")[$frac(3 x + 6, x − 6) < 0$]], [],
  [Factor the numerator to show all factors.], [#math.equation(block: false, alt: "the fraction 3 open parenthesis x plus 2 close parenthesis over x minus 6 less than 0")[$frac(3 ( x + 2 ), x − 6) < 0$]], [],
  [Find the zero partition numbers.], [], [],
  [The quotient will be zero when the numerator is zero. #linebreak() The quotient is undefined when the denominator is zero.], [#math.equation(block: false, alt: "x plus 2, equals, 0, x minus 6, equals, 0; x, equals, − 2, x, equals, 6")[$x + 2 & = & 0 &  &  & x − 6 & = & 0 \ x & = & "−" 2 &  &  & x & = & 6$]], [],
  [Use the zero partition numbers to divide the number line into intervals.], [], [],
  [#figure(figph[A number line with vertical dashed lines at x=-2 and x=6, indicating boundaries or vertical asymptotes.], alt: "A number line with vertical dashed lines at x=-2 and x=6, indicating boundaries or vertical asymptotes.", caption: none)], [], [],
  [Test a value in each interval.], [], [],
  [#figure(figph[A sign analysis table for algebraic expressions x+2 and x-6. It evaluates each expression within intervals (-∞, -2), (-2, 6), and (6, ∞), showing the computed value and its sign.], alt: "A sign analysis table for algebraic expressions x+2 and x-6. It evaluates each expression within intervals (-∞, -2), (-2, 6), and (6, ∞), showing the computed value and its sign.", caption: none)], [], [],
  [Above the number line show the sign of each factor of the rational expression in each interval. #linebreak() Below the number line show the sign of the quotient.], [], [],
  [#figure(figph[A sign analysis chart shows the behavior of the rational expression (x+2)/(x-6) on a number line. It illustrates that the expression is positive for x \< -2 or x \> 6, and negative for -2 \< x \< 6.], alt: "A sign analysis chart shows the behavior of the rational expression (x+2)/(x-6) on a number line. It illustrates that the expression is positive for x < -2 or x > 6, and negative for -2 < x < 6.", caption: none)], [], [],
  [Determine the intervals where the inequality is correct. We want the quotient to be negative, so the solution includes the points between −2 and 6. Since the inequality is strictly less than, the endpoints are not included.], [], [],
  [We write the solution in interval notation as (−2, 6).], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 3 x over x minus 3 less than 1 .")[$frac(3 x, x − 3) < 1 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis minus the fraction 3 over 2 , 3 close parenthesis")[$( − frac(3, 2) , 3 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 3 x over x minus 4 less than 2 .")[$frac(3 x, x − 4) < 2 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −8 , 4 close parenthesis")[$( −8 , 4 )$]
]
]

In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator.

#examplebox("Example 3")[][
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 5 over x squared minus 2 x minus 15 greater than 0 .")[$frac(5, x^(2) − 2 x − 15) > 0 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [The inequality is in the correct form.], [#math.equation(block: false, alt: "the fraction 5 over x squared minus 2 x minus 15 greater than 0")[$frac(5, x^(2) − 2 x − 15) > 0$]],
  [Factor the denominator.], [#math.equation(block: false, alt: "the fraction 5 over open parenthesis x plus 3 close parenthesis open parenthesis x minus 5 close parenthesis greater than 0")[$frac(5, ( x + 3 ) ( x − 5 )) > 0$]],
  [Find the zero partition numbers. #linebreak() The quotient is 0 when the numerator is 0. #linebreak() Since the numerator is always 5, the quotient cannot be 0.], [],
  [The quotient will be undefined when the #linebreak() denominator is zero.], [#math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis open parenthesis x minus 5 close parenthesis equals 0; x equals minus 3 , x equals 5")[$( x + 3 ) ( x − 5 ) = 0 \ x = − 3 , " " x = 5$]],
  [Use the zero partition numbers to divide the number line into intervals.], [],
  [], [#figure(figph[Sign chart showing the intervals where 5/((x+3)(x-5)) is positive or negative. It details the signs of (x+3) and (x-5) across a number line with critical points at -3 and 5.], alt: "Sign chart showing the intervals where 5/((x+3)(x-5)) is positive or negative. It details the signs of (x+3) and (x-5) across a number line with critical points at -3 and 5.", caption: none)],
  [Test values in each interval. #linebreak() Above the number line show the sign of each #linebreak() factor of the denominator in each interval. #linebreak() Below the number line, show the sign of the quotient.], [],
  [Write the solution in interval notation.], [#math.equation(block: false, alt: "open parenthesis − infinity , minus 3 close parenthesis union open parenthesis 5 , infinity close parenthesis")[$( "−" ∞ , − 3 ) ∪ ( 5 , ∞ )$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 1 over x squared plus 2 x minus 8 greater than 0 .")[$frac(1, x^(2) + 2 x − 8) > 0 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 close parenthesis union open parenthesis 2 , ∞ close parenthesis")[$( "−" ∞ , −4 ) ∪ ( 2 , ∞ )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 3 over x squared plus x minus 12 greater than 0 .")[$frac(3, x^(2) + x − 12) > 0 .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 close parenthesis union open parenthesis 3 , ∞ close parenthesis")[$( "−" ∞ , −4 ) ∪ ( 3 , ∞ )$]
]
]

The next example requires some work to get it into the needed form.

#examplebox("Example 4")[][
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 1 over 3 minus the fraction 2 over x squared less than the fraction 5 over 3 x .")[$frac(1, 3) − frac(2, x^(2)) < frac(5, 3 x) .$]

#solutionbox[
#figure(table(
  columns: 3, align: left, inset: 6pt,
  [], [#math.equation(block: false, alt: "the fraction 1 over 3 minus the fraction 2 over x squared less than the fraction 5 over 3 x")[$frac(1, 3) − frac(2, x^(2)) < frac(5, 3 x)$]], [],
  [Subtract #math.equation(block: false, alt: "the fraction 5 over 3 x")[$frac(5, 3 x)$] to get zero on the right.], [#math.equation(block: false, alt: "the fraction 1 over 3 minus the fraction 2 over x squared minus the fraction 5 over 3 x less than 0")[$frac(1, 3) − frac(2, x^(2)) − frac(5, 3 x) < 0$]], [],
  [Rewrite to get each fraction with the LCD #math.equation(block: false, alt: "3 x squared .")[$3 x^(2) .$]], [#math.equation(block: false, alt: "the fraction 1 times x squared over 3 times x squared minus the fraction 2 times 3 over x squared times 3 minus the fraction 5 times x over 3 x times x less than 0")[$frac(1 ⋅ x^(2), 3 ⋅ x^(2)) − frac(2 ⋅ 3, x^(2) ⋅ 3) − frac(5 ⋅ x, 3 x ⋅ x) < 0$]], [],
  [Simplify.], [#math.equation(block: false, alt: "the fraction x squared over 3 x squared minus the fraction 6 over 3 x squared minus the fraction 5 x over 3 x squared less than 0")[$frac(x^(2), 3 x^(2)) − frac(6, 3 x^(2)) − frac(5 x, 3 x^(2)) < 0$]], [],
  [Subtract the numerators and place the #linebreak() difference over the common denominator.], [#math.equation(block: false, alt: "the fraction x squared minus 5 x minus 6 over 3 x squared less than 0")[$frac(x^(2) − 5 x − 6, 3 x^(2)) < 0$]], [],
  [Factor the numerator.], [#math.equation(block: false, alt: "the fraction open parenthesis x minus 6 close parenthesis open parenthesis x plus 1 close parenthesis over 3 x squared less than 0")[$frac(( x − 6 ) ( x + 1 ), 3 x^(2)) < 0$]], [],
  [Find the zero partition numbers.], [#math.equation(block: false, alt: "3 x squared, equals, 0, x minus 6, equals, 0, x plus 1, equals, 0; x, equals, 0, x, equals, 6, x, equals, − 1")[$3 x^(2) & = & 0 &  &  & x − 6 & = & 0 &  &  & x + 1 & = & 0 \ x & = & 0 &  &  & x & = & 6 &  &  & x & = & "−" 1$]], [],
  [Use the zero partition numbers to divide the number #linebreak() line into intervals.], [], [],
  [], [#figure(figph[This image displays a sign chart for the expression (x-6)(x+1) / (3x^2), showing the signs of its factors and the overall expression across intervals defined by critical points -1, 0, and 6.], alt: "This image displays a sign chart for the expression (x-6)(x+1) / (3x^2), showing the signs of its factors and the overall expression across intervals defined by critical points -1, 0, and 6.", caption: none)], [],
  [Above the number line show the sign of each #linebreak() factor in each interval.  Below the number line, show the sign of the quotient.], [], [],
  [Since, 0 is excluded, the solution is the two #linebreak() intervals, #math.equation(block: false, alt: "open parenthesis −1 , 0 close parenthesis")[$( −1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 0 , 6 close parenthesis .")[$( 0 , 6 ) .$]], [#math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis union open parenthesis 0 , 6 close parenthesis")[$( − 1 , 0 ) ∪ ( 0 , 6 )$]], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 1 over 2 plus the fraction 4 over x squared less than the fraction 3 over x .")[$frac(1, 2) + frac(4, x^(2)) < frac(3, x) .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 2 , 4 close parenthesis")[$( 2 , 4 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 1 over 3 plus the fraction 6 over x squared less than the fraction 3 over x .")[$frac(1, 3) + frac(6, x^(2)) < frac(3, x) .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 3 , 6 close parenthesis")[$( 3 , 6 )$]
]
]


=== Solve an Inequality with Rational Functions

When working with rational functions, it is sometimes useful to know when the function is greater than or less than a particular value. This leads to a rational inequality.

#examplebox("Example 5")[][
Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x plus 3 over x minus 5 ,")[$R ( x ) = frac(x + 3, x − 5) ,$] find the values of #emph[x] that make the function less than or equal to 0.

#solutionbox[
We want the function to be less than or equal to 0.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#math.equation(block: false, alt: "R open parenthesis x close parenthesis less than or equal to 0")[$R ( x ) ≤ 0$]],
  [Substitute the rational expression for #math.equation(block: false, alt: "R open parenthesis x close parenthesis .")[$R ( x ) .$]], [#math.equation(block: false, alt: "the fraction x plus 3 over x minus 5 less than or equal to 0 x not equal to 5")[$frac(x + 3, x − 5) ≤ 0 #h(3.5em) x ≠ 5$]],
  [Find the zero partition numbers.], [#math.equation(block: false, alt: "x plus 3, equals, 0, x minus 5, equals, 0; x, equals, −3, x, equals, 5")[$x + 3 & = & 0 &  &  & x − 5 & = & 0 \ x & = & −3 &  &  & x & = & 5$]],
  [Use the zero partition numbers to divide the number line into intervals.], [],
  [], [#figure(figph[A sign chart illustrating the intervals where the rational expression (x+3)/(x-5) is positive or negative, based on the signs of its numerator and denominator.], alt: "A sign chart illustrating the intervals where the rational expression (x+3)/(x-5) is positive or negative, based on the signs of its numerator and denominator.", caption: none)],
  [Test values in each interval. Above the #linebreak() number line, show the sign of each factor #linebreak() in each interval. Below the number line, #linebreak() show the sign of the quotient], [],
  [Write the solution in interval notation. Since #linebreak() 5 is excluded we, do not include it in the interval.], [#math.equation(block: false, alt: "[ −3 , 5 close parenthesis")[$[ −3 , 5 )$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x minus 2 over x plus 4 ,")[$R ( x ) = frac(x − 2, x + 4) ,$] find the values of #emph[x] that make the function less than or equal to 0.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −4 , 2 ]")[$( −4 , 2 ]$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x plus 1 over x minus 4 ,")[$R ( x ) = frac(x + 1, x − 4) ,$] find the values of #emph[x] that make the function less than or equal to 0.

#solutionbox[
#math.equation(block: true, alt: "[ −1 , 4 close parenthesis")[$[ −1 , 4 )$]
]
]

In economics, the function #math.equation(block: false, alt: "C open parenthesis x close parenthesis")[$C ( x )$] is used to represent the cost of producing #emph[x] units of a commodity. The average cost per unit can be found by dividing #math.equation(block: false, alt: "C open parenthesis x close parenthesis")[$C ( x )$] by the number of items #math.equation(block: false, alt: "x .")[$x .$] Then, the average cost per unit is #math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction C open parenthesis x close parenthesis over x .")[$c ( x ) = frac(C ( x ), x) .$]

#examplebox("Example 6")[][
The function #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 10 x plus 3000")[$C ( x ) = 10 x + 3000$] represents the cost to produce #math.equation(block: false, alt: "x ,")[$x ,$] number of items. Find ⓐ the average cost function, #math.equation(block: false, alt: "c open parenthesis x close parenthesis")[$c ( x )$] ⓑ how many items should be produced so that the average cost is less than \$40.

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 10 x plus 3000")[$C ( x ) = 10 x + 3000$]]),
  [The average cost function is #math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction C open parenthesis x close parenthesis over x .")[$c ( x ) = frac(C ( x ), x) .$]], [],
  [To find the average cost function, divide the #linebreak() cost function by #math.equation(block: false, alt: "x .")[$x .$]], [#math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction C open parenthesis x close parenthesis over x; c open parenthesis x close parenthesis equals the fraction 10 x plus 3000 over x")[$c ( x ) = frac(C ( x ), x) \ c ( x ) = frac(10 x + 3000, x)$]],
  [], [The average cost function is #math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction 10 x plus 3000 over x .")[$c ( x ) = frac(10 x + 3000, x) .$]],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We want the function #math.equation(block: false, alt: "c open parenthesis x close parenthesis")[$c ( x )$] to be less than #math.equation(block: false, alt: "40 .")[$40 .$]], [#math.equation(block: false, alt: "c open parenthesis x close parenthesis less than 40")[$#h(6.4em) c ( x ) < 40$]]),
  [Substitute the rational expression for #math.equation(block: false, alt: "c open parenthesis x close parenthesis .")[$c ( x ) .$]], [#math.equation(block: false, alt: "the fraction 10 x plus 3000 over x less than 40 x not equal to 0")[$#h(3.3em) frac(10 x + 3000, x) < 40 #h(0.5em) x ≠ 0$]],
  [Subtract 40 to get 0 on the right.], [#math.equation(block: false, alt: "the fraction 10 x plus 3000 over x minus 40 less than 0")[$#h(1.11em) frac(10 x + 3000, x) − 40 < 0$]],
  [Rewrite the left side as one quotient by finding #linebreak() the LCD and performing the subtraction.], [#math.equation(block: false, alt: "the fraction 10 x plus 3000 over x minus 40 open parenthesis the fraction x over x close parenthesis less than 0")[$frac(10 x + 3000, x) − 40 ( frac(x, x) ) < 0$]],
  [], [#math.equation(block: false, alt: "the fraction 10 x plus 3000 over x minus the fraction 40 x over x less than 0")[$#h(0.5em) frac(10 x + 3000, x) − frac(40 x, x) < 0$]],
  [], [#math.equation(block: false, alt: "the fraction 10 x plus 3000 minus 40 x over x less than 0")[$#h(0.6em) frac(10 x + 3000 − 40 x, x) < 0$]],
  [], [#math.equation(block: false, alt: "the fraction −30 x plus 3000 over x less than 0")[$#h(2.55em) frac(−30 x + 3000, x) < 0$]],
  [Factor the numerator to show all factors.], [#math.equation(block: false, alt: "the fraction −30 open parenthesis x minus 100 close parenthesis over x less than 0")[$#h(2.4em) frac(−30 ( x − 100 ), x) < 0$]],
  [Find the zero partition numbers.], [#math.equation(block: false, alt: "minus 30 open parenthesis x minus 100 close parenthesis, equals, 0, x equals 0; minus 30 not equal to 0 x minus 100, equals, 0; x, equals, 100")[$− 30 ( x − 100 ) & = & 0 & #h(1em) x = 0 \ − 30 ≠ 0 #h(1em) x − 100 & = & 0 &  \ x & = & 100 &$]],
))

More than 100 items must be produced to keep the average cost below \$40 per item.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The function #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 20 x plus 6000")[$C ( x ) = 20 x + 6000$] represents the cost to produce #math.equation(block: false, alt: "x ,")[$x ,$] number of items. Find ⓐ the average cost function, #math.equation(block: false, alt: "c open parenthesis x close parenthesis")[$c ( x )$] ⓑ how many items should be produced so that the average cost is less than \$60?

#solutionbox[
ⓐ #math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction 20 x plus 6000 over x")[$c ( x ) = frac(20 x + 6000, x)$] #linebreak() ⓑ More than 150 items must be produced to keep the average cost below \$60 per item.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The function #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 5 x plus 900")[$C ( x ) = 5 x + 900$] represents the cost to produce #math.equation(block: false, alt: "x ,")[$x ,$] number of items. Find ⓐ the average cost function, #math.equation(block: false, alt: "c open parenthesis x close parenthesis")[$c ( x )$] ⓑ how many items should be produced so that the average cost is less than \$20?

#solutionbox[
ⓐ #math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction 5 x plus 900 over x")[$c ( x ) = frac(5 x + 900, x)$] ⓑ More than 60 items must be produced to keep the average cost below \$20 per item.
]
]


=== Key Concepts

- #strong[Solve a rational inequality.]
+ Write the inequality as one quotient on the left and zero on the right.
+ Determine the zero partition numbers–the points where the rational expression will be zero or undefined.
+ Use the zero partition numbers to divide the number line into intervals.
+ Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.
+ Determine the intervals where the inequality is correct. Write the solution in interval notation.


=== Section Exercises


==== Practice Makes Perfect

#strong[Solve Rational Inequalities]

In the following exercises, solve each rational inequality and write the solution in interval notation.

#math.equation(block: true, alt: "the fraction x minus 3 over x plus 4 greater than or equal to 0")[$frac(x − 3, x + 4) ≥ 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 close parenthesis union [ 3 , ∞ close parenthesis")[$( "−" ∞ , −4 ) ∪ [ 3 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction x plus 6 over x minus 5 greater than or equal to 0")[$frac(x + 6, x − 5) ≥ 0$]

#math.equation(block: true, alt: "the fraction x plus 1 over x minus 3 less than or equal to 0")[$frac(x + 1, x − 3) ≤ 0$]

#solutionbox[
#math.equation(block: true, alt: "[ −1 , 3 close parenthesis")[$[ −1 , 3 )$]
]

#math.equation(block: true, alt: "the fraction x minus 4 over x plus 2 less than or equal to 0")[$frac(x − 4, x + 2) ≤ 0$]

#math.equation(block: true, alt: "the fraction x minus 7 over x minus 1 greater than 0")[$frac(x − 7, x − 1) > 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , 1 close parenthesis union open parenthesis 7 , ∞ close parenthesis")[$( "−" ∞ , 1 ) ∪ ( 7 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction x plus 8 over x plus 3 greater than 0")[$frac(x + 8, x + 3) > 0$]

#math.equation(block: true, alt: "the fraction x minus 6 over x plus 5 less than 0")[$frac(x − 6, x + 5) < 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −5 , 6 close parenthesis")[$( −5 , 6 )$]
]

#math.equation(block: true, alt: "the fraction x plus 5 over x minus 2 less than 0")[$frac(x + 5, x − 2) < 0$]

#math.equation(block: true, alt: "the fraction 3 x over x minus 5 less than 1")[$frac(3 x, x − 5) < 1$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis minus the fraction 5 over 2 , 5 close parenthesis")[$( − frac(5, 2) , 5 )$]
]

#math.equation(block: true, alt: "the fraction 5 x over x minus 2 less than 1")[$frac(5 x, x − 2) < 1$]

#math.equation(block: true, alt: "the fraction 6 x over x minus 6 greater than 2")[$frac(6 x, x − 6) > 2$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −3 close parenthesis union open parenthesis 6 , ∞ close parenthesis")[$( "−" ∞ , −3 ) ∪ ( 6 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 3 x over x minus 4 greater than 2")[$frac(3 x, x − 4) > 2$]

#math.equation(block: true, alt: "the fraction 2 x plus 3 over x minus 6 less than or equal to 1")[$frac(2 x + 3, x − 6) ≤ 1$]

#solutionbox[
#math.equation(block: true, alt: "[ −9 , 6 close parenthesis")[$[ −9 , 6 )$]
]

#math.equation(block: true, alt: "the fraction 4 x minus 1 over x minus 4 less than or equal to 1")[$frac(4 x − 1, x − 4) ≤ 1$]

#math.equation(block: true, alt: "the fraction 3 x minus 2 over x minus 4 greater than or equal to 2")[$frac(3 x − 2, x − 4) ≥ 2$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −6 ] union open parenthesis 4 , ∞ close parenthesis")[$( "−" ∞ , −6 ] ∪ ( 4 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 4 x minus 3 over x minus 3 greater than or equal to 2")[$frac(4 x − 3, x − 3) ≥ 2$]

#math.equation(block: true, alt: "the fraction 1 over x squared plus 7 x plus 12 greater than 0")[$frac(1, x^(2) + 7 x + 12) > 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 close parenthesis union open parenthesis −3 , ∞ close parenthesis")[$( "−" ∞ , −4 ) ∪ ( −3 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 1 over x squared minus 4 x minus 12 greater than 0")[$frac(1, x^(2) − 4 x − 12) > 0$]

#math.equation(block: true, alt: "the fraction 3 over x squared minus 5 x plus 4 less than 0")[$frac(3, x^(2) − 5 x + 4) < 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$]
]

#math.equation(block: true, alt: "the fraction 4 over x squared plus 7 x plus 12 less than 0")[$frac(4, x^(2) + 7 x + 12) < 0$]

#math.equation(block: true, alt: "the fraction 2 over 2 x squared plus x minus 15 greater than or equal to 0")[$frac(2, 2 x^(2) + x − 15) ≥ 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −3 close parenthesis union open parenthesis the fraction 5 over 2 , ∞ close parenthesis")[$( "−" ∞ , −3 ) ∪ ( frac(5, 2) , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 6 over 3 x squared minus 2 x minus 5 greater than or equal to 0")[$frac(6, 3 x^(2) − 2 x − 5) ≥ 0$]

#math.equation(block: true, alt: "the fraction −2 over 6 x squared minus 13 x plus 6 less than or equal to 0")[$frac(−2, 6 x^(2) − 13 x + 6) ≤ 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , the fraction 2 over 3 close parenthesis union open parenthesis the fraction 3 over 2 , ∞ close parenthesis")[$( "−" ∞ , frac(2, 3) ) ∪ ( frac(3, 2) , ∞ )$]
]

#math.equation(block: true, alt: "the fraction −1 over 10 x squared plus 11 x minus 6 less than or equal to 0")[$frac(−1, 10 x^(2) + 11 x − 6) ≤ 0$]

#math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 12 over x squared greater than the fraction 5 over x")[$frac(1, 2) + frac(12, x^(2)) > frac(5, x)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , 0 close parenthesis union open parenthesis 0 , 4 close parenthesis union open parenthesis 6 , ∞ close parenthesis")[$( "−" ∞ , 0 ) ∪ ( 0 , 4 ) ∪ ( 6 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 1 over 3 plus the fraction 1 over x squared greater than the fraction 4 over 3 x")[$frac(1, 3) + frac(1, x^(2)) > frac(4, 3 x)$]

#math.equation(block: true, alt: "the fraction 1 over 2 minus the fraction 4 over x squared less than or equal to the fraction 1 over x")[$frac(1, 2) − frac(4, x^(2)) ≤ frac(1, x)$]

#solutionbox[
#math.equation(block: true, alt: "[ −2 , 0 close parenthesis union open parenthesis 0 , 4 ]")[$[ −2 , 0 ) ∪ ( 0 , 4 ]$]
]

#math.equation(block: true, alt: "the fraction 1 over 2 minus the fraction 3 over 2 x squared greater than or equal to the fraction 1 over x")[$frac(1, 2) − frac(3, 2 x^(2)) ≥ frac(1, x)$]

#math.equation(block: true, alt: "the fraction 1 over x squared minus 16 less than 0")[$frac(1, x^(2) − 16) < 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −4 , 4 close parenthesis")[$( −4 , 4 )$]
]

#math.equation(block: true, alt: "the fraction 4 over x squared minus 25 greater than 0")[$frac(4, x^(2) − 25) > 0$]

#math.equation(block: true, alt: "the fraction 4 over x minus 2 greater than or equal to the fraction 3 over x plus 1")[$frac(4, x − 2) ≥ frac(3, x + 1)$]

#solutionbox[
#math.equation(block: true, alt: "[ −10 , −1 close parenthesis union open parenthesis 2 , ∞ close parenthesis")[$[ −10 , −1 ) ∪ ( 2 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 5 over x minus 1 less than or equal to the fraction 4 over x plus 2")[$frac(5, x − 1) ≤ frac(4, x + 2)$]

#strong[Solve an Inequality with Rational Functions]

In the following exercises, solve each rational function inequality and write the solution in interval notation.

Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x minus 5 over x minus 2 ,")[$R ( x ) = frac(x − 5, x − 2) ,$] find the values of #math.equation(block: false, alt: "x")[$x$] that make the function less than or equal to 0.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 2 , 5 ]")[$( 2 , 5 ]$]
]

Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x plus 1 over x plus 3 ,")[$R ( x ) = frac(x + 1, x + 3) ,$] find the values of #math.equation(block: false, alt: "x")[$x$] that make the function greater than or equal to 0.

Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x minus 6 over x plus 2")[$R ( x ) = frac(x − 6, x + 2)$], find the values of x that make the function less than or equal to 0.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −2 , 6 ]")[$( −2 , 6 ]$]
]

Given the function #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x plus 1 over x minus 4 ,")[$R ( x ) = frac(x + 1, x − 4) ,$] find the values of x that make the function less than or equal to 0.


==== Writing Exercises

Write the steps you would use to explain solving rational inequalities to your little brother.

#solutionbox[
Answers will vary.
]

Create a rational inequality whose solution is #math.equation(block: false, alt: "open parenthesis − ∞ , −2 ] union [ 4 , ∞ close parenthesis .")[$( "−" ∞ , −2 ] ∪ [ 4 , ∞ ) .$]


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve rational inequalities. In row 3, the I can was solve an inequality with rational functions.], alt: "This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve rational inequalities. In row 3, the I can was solve an inequality with rational functions.", caption: none)

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?


=== Chapter Review Exercises


==== Simplify, Multiply, and Divide Rational Expressions

#strong[Determine the Values for Which a Rational Expression is Undefined]

In the following exercises, determine the values for which the rational expression is undefined.

#math.equation(block: true, alt: "the fraction 5 a plus 3 over 3 a minus 2")[$frac(5 a + 3, 3 a − 2)$]

#solutionbox[
#math.equation(block: true, alt: "a not equal to the fraction 2 over 3")[$a ≠ frac(2, 3)$]
]

#math.equation(block: true, alt: "the fraction b minus 7 over b squared minus 25")[$frac(b − 7, b^(2) − 25)$]

#math.equation(block: true, alt: "the fraction 5 x squared y squared over 8 y")[$frac(5 x^(2) y^(2), 8 y)$]

#solutionbox[
#math.equation(block: true, alt: "y not equal to 0")[$y ≠ 0$]
]

#math.equation(block: true, alt: "the fraction x minus 3 over x squared minus x minus 30")[$frac(x − 3, x^(2) − x − 30)$]

#strong[Simplify Rational Expressions]

In the following exercises, simplify.

#math.equation(block: true, alt: "the fraction 18 over 24")[$frac(18, 24)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 3 over 4")[$frac(3, 4)$]
]

#math.equation(block: true, alt: "the fraction 9 m to the power 4 over 18 m n cubed")[$frac(9 m^(4), 18 m n^(3))$]

#math.equation(block: true, alt: "the fraction x squared plus 7 x plus 12 over x squared plus 8 x plus 16")[$frac(x^(2) + 7 x + 12, x^(2) + 8 x + 16)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction x plus 3 over x plus 4")[$frac(x + 3, x + 4)$]
]

#math.equation(block: true, alt: "the fraction 7 v minus 35 over 25 minus v squared")[$frac(7 v − 35, 25 − v^(2))$]

#strong[Multiply Rational Expressions]

In the following exercises, multiply.

#math.equation(block: true, alt: "the fraction 5 over 8 times the fraction 4 over 15")[$frac(5, 8) · frac(4, 15)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 1 over 6")[$frac(1, 6)$]
]

#math.equation(block: true, alt: "the fraction 3 x y squared over 8 y cubed times the fraction 16 y squared over 24 x")[$frac(3 x y^(2), 8 y^(3)) · frac(16 y^(2), 24 x)$]

#math.equation(block: true, alt: "the fraction 72 x minus 12 x squared over 8 x plus 32 times the fraction x squared plus 10 x plus 24 over x squared minus 36")[$frac(72 x − 12 x^(2), 8 x + 32) · frac(x^(2) + 10 x + 24, x^(2) − 36)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction −3 x over 2")[$frac(−3 x, 2)$]
]

#math.equation(block: true, alt: "the fraction 2 y squared plus y minus 3 over 4 minus y squared times the fraction y squared minus 4 y plus 4 over 2 y squared plus 11 y plus 12")[$frac(2 y^(2) + y − 3, 4 − y^(2)) · frac(y^(2) − 4 y + 4, 2 y^(2) + 11 y + 12)$]

#strong[Divide Rational Expressions]

In the following exercises, divide.

#math.equation(block: true, alt: "the fraction x squared minus 4 x minus 12 over x squared plus 8 x plus 12 divided by the fraction x squared minus 36 over 3 x")[$frac(x^(2) − 4 x − 12, x^(2) + 8 x + 12) ÷ frac(x^(2) − 36, 3 x)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 3 x over open parenthesis x plus 6 close parenthesis open parenthesis x plus 6 close parenthesis")[$frac(3 x, ( x + 6 ) ( x + 6 ))$]
]

#math.equation(block: true, alt: "the fraction y squared minus 16 over 4 divided by the fraction y cubed minus 64 over 2 y squared plus 8 y plus 32")[$frac(y^(2) − 16, 4) ÷ frac(y^(3) − 64, 2 y^(2) + 8 y + 32)$]

#math.equation(block: true, alt: "the fraction 11 plus w over w minus 9 divided by the fraction 121 minus w squared over 9 minus w")[$frac(11 + w, w − 9) ÷ frac(121 − w^(2), 9 − w)$]

#solutionbox[
#math.equation(block: true, alt: "minus the fraction 1 over 11 minus w")[$− frac(1, 11 − w)$]
]

#math.equation(block: true, alt: "the fraction 3 y squared minus 12 y minus 63 over 4 y plus 3 divided by open parenthesis 6 y squared minus 42 y close parenthesis")[$frac(3 y^(2) − 12 y − 63, 4 y + 3) ÷ ( 6 y^(2) − 42 y )$]

#math.equation(block: true, alt: "the fraction the fraction c squared minus 64 over 3 c squared plus 26 c plus 16 over the fraction c squared minus 4 c minus 32 over 15 c plus 10")[$frac(frac(c^(2) − 64, 3 c^(2) + 26 c + 16), frac(c^(2) − 4 c − 32, 15 c + 10))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 5 over c plus 4")[$frac(5, c + 4)$]
]

#math.equation(block: true, alt: "the fraction 8 a squared plus 16 a over a minus 4 times the fraction a squared plus 2 a minus 24 over a squared plus 7 a plus 10 divided by the fraction 2 a squared minus 6 a over a plus 5")[$frac(8 a^(2) + 16 a, a − 4) · frac(a^(2) + 2 a − 24, a^(2) + 7 a + 10) ÷ frac(2 a^(2) − 6 a, a + 5)$]

#strong[Multiply and Divide Rational Functions]

Find #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals f open parenthesis x close parenthesis times g open parenthesis x close parenthesis")[$R ( x ) = f ( x ) · g ( x )$] where #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 9 x squared plus 9 x over x squared minus 3 x minus 4")[$f ( x ) = frac(9 x^(2) + 9 x, x^(2) − 3 x − 4)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x squared minus 16 over 3 x squared plus 12 x .")[$g ( x ) = frac(x^(2) − 16, 3 x^(2) + 12 x) .$]

#solutionbox[
#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals 3")[$R ( x ) = 3$]
]

Find #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction f open parenthesis x close parenthesis over g open parenthesis x close parenthesis")[$R ( x ) = frac(f ( x ), g ( x ))$] where #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 27 x squared over 3 x minus 21")[$f ( x ) = frac(27 x^(2), 3 x − 21)$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 9 x squared plus 54 x over x squared minus x minus 42 .")[$g ( x ) = frac(9 x^(2) + 54 x, x^(2) − x − 42) .$]


==== Add and Subtract Rational Expressions

#strong[Add and Subtract Rational Expressions with a Common Denominator]

In the following exercises, perform the indicated operations.

#math.equation(block: true, alt: "the fraction 7 over 15 plus the fraction 8 over 15")[$frac(7, 15) + frac(8, 15)$]

#solutionbox[
#math.equation(block: true, alt: "1")[$1$]
]

#math.equation(block: true, alt: "the fraction 4 a squared over 2 a minus 1 minus the fraction 1 over 2 a minus 1")[$frac(4 a^(2), 2 a − 1) − frac(1, 2 a − 1)$]

#math.equation(block: true, alt: "the fraction y squared plus 10 y over y plus 5 plus the fraction 25 over y plus 5")[$frac(y^(2) + 10 y, y + 5) + frac(25, y + 5)$]

#solutionbox[
#math.equation(block: true, alt: "y plus 5")[$y + 5$]
]

#math.equation(block: true, alt: "the fraction 7 x squared over x squared minus 9 plus the fraction 21 x over x squared minus 9")[$frac(7 x^(2), x^(2) − 9) + frac(21 x, x^(2) − 9)$]

#math.equation(block: true, alt: "the fraction x squared over x minus 7 minus the fraction 3 x plus 28 over x minus 7")[$frac(x^(2), x − 7) − frac(3 x + 28, x − 7)$]

#solutionbox[
#math.equation(block: true, alt: "x plus 4")[$x + 4$]
]

#math.equation(block: true, alt: "the fraction y squared over y plus 11 minus the fraction 121 over y plus 11")[$frac(y^(2), y + 11) − frac(121, y + 11)$]

#math.equation(block: true, alt: "the fraction 4 q squared minus q plus 3 over q squared plus 6 q plus 5 minus the fraction 3 q squared plus q plus 6 over q squared plus 6 q plus 5")[$frac(4 q^(2) − q + 3, q^(2) + 6 q + 5) − frac(3 q^(2) + q + 6, q^(2) + 6 q + 5)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction q minus 3 over q plus 5")[$frac(q − 3, q + 5)$]
]

#math.equation(block: true, alt: "the fraction 5 t squared plus 4 t squared plus 3 over t squared minus 25 minus the fraction 4 t squared minus 8 t minus 32 over t squared minus 25")[$frac(5 t^(2) + 4 t^(2) + 3, t^(2) − 25) − frac(4 t^(2) − 8 t − 32, t^(2) − 25)$]

#strong[Add and Subtract Rational Expressions Whose Denominators Are Opposites]

In the following exercises, add and subtract.

#math.equation(block: true, alt: "the fraction 18 w over 6 w minus 1 plus the fraction 3 w minus 2 over 1 minus 6 w")[$frac(18 w, 6 w − 1) + frac(3 w − 2, 1 − 6 w)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 15 w plus 2 over 6 w minus 1")[$frac(15 w + 2, 6 w − 1)$]
]

#math.equation(block: true, alt: "the fraction a squared plus 3 a over a squared minus 4 minus the fraction 3 a plus 8 over 4 minus a squared")[$frac(a^(2) + 3 a, a^(2) − 4) − frac(3 a + 8, 4 − a^(2))$]

#math.equation(block: true, alt: "the fraction 2 b squared plus 3 b minus 15 over b squared minus 49 minus the fraction b squared plus 16 b minus 1 over 49 minus b squared")[$frac(2 b^(2) + 3 b − 15, b^(2) − 49) − frac(b^(2) + 16 b − 1, 49 − b^(2))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 3 b squared plus 19 b minus 16 over b squared minus 49")[$frac(3 b^(2) + 19 b − 16, b^(2) − 49)$]
]

#math.equation(block: true, alt: "the fraction 8 y squared minus 10 y plus 7 over 2 y minus 5 plus the fraction 2 y squared plus 7 y plus 2 over 5 minus 2 y")[$frac(8 y^(2) − 10 y + 7, 2 y − 5) + frac(2 y^(2) + 7 y + 2, 5 − 2 y)$]

#strong[Find the Least Common Denominator of Rational Expressions]

In the following exercises, find the LCD.

#math.equation(block: true, alt: "the fraction 7 over a squared minus 3 a minus 10 , the fraction 3 a over a squared minus a minus 20")[$frac(7, a^(2) − 3 a − 10) , frac(3 a, a^(2) − a − 20)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a plus 2 close parenthesis open parenthesis a minus 5 close parenthesis open parenthesis a plus 4 close parenthesis")[$( a + 2 ) ( a − 5 ) ( a + 4 )$]
]

#math.equation(block: true, alt: "the fraction 6 over n squared minus 4 , the fraction 2 n over n squared minus 4 n plus 4")[$frac(6, n^(2) − 4) , frac(2 n, n^(2) − 4 n + 4)$]

#math.equation(block: true, alt: "the fraction 5 over 3 p squared plus 17 p minus 6 , the fraction 2 m over 3 p squared plus 23 p minus 8")[$frac(5, 3 p^(2) + 17 p − 6) , frac(2 m, 3 p^(2) + 23 p − 8)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 3 p minus 1 close parenthesis open parenthesis p plus 6 close parenthesis open parenthesis p plus 8 close parenthesis")[$( 3 p − 1 ) ( p + 6 ) ( p + 8 )$]
]

#strong[Add and Subtract Rational Expressions with Unlike Denominators]

In the following exercises, perform the indicated operations.

#math.equation(block: true, alt: "the fraction 7 over 5 a plus the fraction 3 over 2 b")[$frac(7, 5 a) + frac(3, 2 b)$]

#math.equation(block: true, alt: "the fraction 2 over c minus 2 plus the fraction 9 over c plus 3")[$frac(2, c − 2) + frac(9, c + 3)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 11 c minus 12 over open parenthesis c minus 2 close parenthesis open parenthesis c plus 3 close parenthesis")[$frac(11 c − 12, ( c − 2 ) ( c + 3 ))$]
]

#math.equation(block: true, alt: "the fraction 3 x over x squared minus 9 plus the fraction 5 over x squared plus 6 x plus 9")[$frac(3 x, x^(2) − 9) + frac(5, x^(2) + 6 x + 9)$]

#math.equation(block: true, alt: "the fraction 2 x over x squared plus 10 x plus 24 plus the fraction 3 x over x squared plus 8 x plus 16")[$frac(2 x, x^(2) + 10 x + 24) + frac(3 x, x^(2) + 8 x + 16)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 5 x squared plus 26 x over open parenthesis x plus 4 close parenthesis open parenthesis x plus 4 close parenthesis open parenthesis x plus 6 close parenthesis")[$frac(5 x^(2) + 26 x, ( x + 4 ) ( x + 4 ) ( x + 6 ))$]
]

#math.equation(block: true, alt: "the fraction 5 q over p squared q minus p squared plus the fraction 4 q over q squared minus 1")[$frac(5 q, p^(2) q − p^(2)) + frac(4 q, q^(2) − 1)$]

#math.equation(block: true, alt: "the fraction 3 y over y plus 2 minus the fraction y plus 2 over y plus 8")[$frac(3 y, y + 2) − frac(y + 2, y + 8)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 2 open parenthesis y squared plus 10 y minus 2 close parenthesis over open parenthesis y plus 2 close parenthesis open parenthesis y plus 8 close parenthesis")[$frac(2 ( y^(2) + 10 y − 2 ), ( y + 2 ) ( y + 8 ))$]
]

#math.equation(block: true, alt: "the fraction −3 w minus 15 over w squared plus w minus 20 minus the fraction w plus 2 over 4 minus w")[$frac(−3 w − 15, w^(2) + w − 20) − frac(w + 2, 4 − w)$]

#math.equation(block: true, alt: "the fraction 7 m plus 3 over m plus 2 minus 5")[$frac(7 m + 3, m + 2) − 5$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 2 m minus 7 over m plus 2")[$frac(2 m − 7, m + 2)$]
]

#math.equation(block: true, alt: "the fraction n over n plus 3 plus the fraction 2 over n minus 3 minus the fraction n plus 9 over n squared minus 9")[$frac(n, n + 3) + frac(2, n − 3) − frac(n + 9, n^(2) − 9)$]

#math.equation(block: true, alt: "the fraction 8 a over a squared minus 64 minus the fraction 4 over a plus 8")[$frac(8 a, a^(2) − 64) − frac(4, a + 8)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 4 over a minus 8")[$frac(4, a − 8)$]
]

#math.equation(block: true, alt: "the fraction 5 over 12 x squared y plus the fraction 7 over 20 x y cubed")[$frac(5, 12 x^(2) y) + frac(7, 20 x y^(3))$]

#strong[Add and Subtract Rational Functions]

In the following exercises, find #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals f open parenthesis x close parenthesis plus g open parenthesis x close parenthesis")[$R ( x ) = f ( x ) + g ( x )$] where #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] are given.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 2 x squared plus 12 x minus 11 over x squared plus 3 x minus 10 ,")[$f ( x ) = frac(2 x^(2) + 12 x − 11, x^(2) + 3 x − 10) ,$]#math.equation(block: true, alt: "g open parenthesis x close parenthesis equals the fraction x plus 1 over 2 minus x")[$g ( x ) = frac(x + 1, 2 − x)$]

#solutionbox[
#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals the fraction x plus 8 over x plus 5")[$R ( x ) = frac(x + 8, x + 5)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction −4 x plus 31 over x squared plus x minus 30 ,")[$f ( x ) = frac(−4 x + 31, x^(2) + x − 30) ,$]#math.equation(block: true, alt: "g open parenthesis x close parenthesis equals the fraction 5 over x plus 6")[$g ( x ) = frac(5, x + 6)$]

In the following exercises, find #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals f open parenthesis x close parenthesis minus g open parenthesis x close parenthesis")[$R ( x ) = f ( x ) − g ( x )$] where #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] are given.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 4 x over x squared minus 121 ,")[$f ( x ) = frac(4 x, x^(2) − 121) ,$]#math.equation(block: true, alt: "g open parenthesis x close parenthesis equals the fraction 2 over x minus 11")[$g ( x ) = frac(2, x − 11)$]

#solutionbox[
#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals the fraction 2 over x plus 11")[$R ( x ) = frac(2, x + 11)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 7 over x plus 6 ,")[$f ( x ) = frac(7, x + 6) ,$]#math.equation(block: true, alt: "g open parenthesis x close parenthesis equals the fraction 14 x over x squared minus 36")[$g ( x ) = frac(14 x, x^(2) − 36)$]


==== Simplify Complex Rational Expressions

#strong[Simplify a Complex Rational Expression by Writing It as Division]

In the following exercises, simplify.

#math.equation(block: true, alt: "the fraction the fraction 7 x over x plus 2 over the fraction 14 x squared over x squared minus 4")[$frac(frac(7 x, x + 2), frac(14 x^(2), x^(2) − 4))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction x minus 2 over 2 x")[$frac(x − 2, 2 x)$]
]

#math.equation(block: true, alt: "the fraction the fraction 2 over 5 plus the fraction 5 over 6 over the fraction 1 over 3 plus the fraction 1 over 4")[$frac(frac(2, 5) + frac(5, 6), frac(1, 3) + frac(1, 4))$]

#math.equation(block: true, alt: "the fraction x minus the fraction 3 x over x plus 5 over the fraction 1 over x plus 5 plus the fraction 1 over x minus 5")[$frac(x − frac(3 x, x + 5), frac(1, x + 5) + frac(1, x − 5))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction open parenthesis x plus 2 close parenthesis open parenthesis x minus 5 close parenthesis over 2")[$frac(( x + 2 ) ( x − 5 ), 2)$]
]

#math.equation(block: true, alt: "the fraction the fraction 2 over m plus the fraction m over n over the fraction n over m minus the fraction 1 over n")[$frac(frac(2, m) + frac(m, n), frac(n, m) − frac(1, n))$]

#strong[Simplify a Complex Rational Expression by Using the LCD]

In the following exercises, simplify.

#math.equation(block: true, alt: "the fraction the fraction 1 over 3 plus the fraction 1 over 8 over the fraction 1 over 4 plus the fraction 1 over 12")[$frac(frac(1, 3) + frac(1, 8), frac(1, 4) + frac(1, 12))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 11 over 8")[$frac(11, 8)$]
]

#math.equation(block: true, alt: "the fraction the fraction 3 over a squared minus the fraction 1 over b over the fraction 1 over a plus the fraction 1 over b squared")[$frac(frac(3, a^(2)) − frac(1, b), frac(1, a) + frac(1, b^(2)))$]

#math.equation(block: true, alt: "the fraction the fraction 2 over z squared minus 49 plus the fraction 1 over z plus 7 over the fraction 9 over z plus 7 plus the fraction 12 over z minus 7")[$frac(frac(2, z^(2) − 49) + frac(1, z + 7), frac(9, z + 7) + frac(12, z − 7))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction z minus 5 over 21 z plus 21")[$frac(z − 5, 21 z + 21)$]
]

#math.equation(block: true, alt: "the fraction the fraction 3 over y squared minus 4 y minus 32 over the fraction 2 over y minus 8 plus the fraction 1 over y plus 4")[$frac(frac(3, y^(2) − 4 y − 32), frac(2, y − 8) + frac(1, y + 4))$]


==== 7.4 Solve Rational Equations

#strong[Solve Rational Equations]

In the following exercises, solve.

#math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 2 over 3 equals the fraction 1 over x")[$frac(1, 2) + frac(2, 3) = frac(1, x)$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 6 over 7")[$x = frac(6, 7)$]
]

#math.equation(block: true, alt: "1 minus the fraction 2 over m equals the fraction 8 over m squared")[$1 − frac(2, m) = frac(8, m^(2))$]

#math.equation(block: true, alt: "the fraction 1 over b minus 2 plus the fraction 1 over b plus 2 equals the fraction 3 over b squared minus 4")[$frac(1, b − 2) + frac(1, b + 2) = frac(3, b^(2) − 4)$]

#solutionbox[
#math.equation(block: true, alt: "b equals the fraction 3 over 2")[$b = frac(3, 2)$]
]

#math.equation(block: true, alt: "the fraction 3 over q plus 8 minus the fraction 2 over q minus 2 equals 1")[$frac(3, q + 8) − frac(2, q − 2) = 1$]

#math.equation(block: true, alt: "the fraction v minus 15 over v squared minus 9 v plus 18 equals the fraction 4 over v minus 3 plus the fraction 2 over v minus 6")[$frac(v − 15, v^(2) − 9 v + 18) = frac(4, v − 3) + frac(2, v − 6)$]

#solutionbox[
no solution
]

#math.equation(block: true, alt: "the fraction z over 12 plus the fraction z plus 3 over 3 z equals the fraction 1 over z")[$frac(z, 12) + frac(z + 3, 3 z) = frac(1, z)$]

#strong[Solve Rational Equations that Involve Functions]

For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x plus 2 over x squared minus 6 x plus 8 ,")[$f ( x ) = frac(x + 2, x^(2) − 6 x + 8) ,$] ⓐ find the domain of the function ⓑ solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1")[$f ( x ) = 1$] ⓒ find the points on the graph at this function value.

#solutionbox[
ⓐ The domain is all real numbers except #math.equation(block: false, alt: "x not equal to 2")[$x ≠ 2$] and #math.equation(block: false, alt: "x not equal to 4 .")[$x ≠ 4 .$] ⓑ #math.equation(block: false, alt: "x equals 1 , x equals 6")[$x = 1 , x = 6$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis , open parenthesis 6 , 1 close parenthesis")[$( 1 , 1 ) , ( 6 , 1 )$]
]

For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 minus x over x squared plus 7 x plus 10 ,")[$f ( x ) = frac(2 − x, x^(2) + 7 x + 10) ,$] ⓐ find the domain of the function ⓑ solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2")[$f ( x ) = 2$] ⓒ find the points on the graph at this function value.

#strong[Solve a Rational Equation for a Specific Variable]

In the following exercises, solve for the indicated variable.

#math.equation(block: false, alt: "the fraction V over l equals h w")[$frac(V, l) = h w$] for #math.equation(block: false, alt: "l .")[$l .$]

#solutionbox[
#math.equation(block: true, alt: "l equals the fraction V over h w")[$l = frac(V, h w)$]
]

#math.equation(block: false, alt: "the fraction 1 over x minus the fraction 2 over y equals 5")[$frac(1, x) − frac(2, y) = 5$] for #math.equation(block: false, alt: "y .")[$y .$]

#math.equation(block: false, alt: "x equals the fraction y plus 5 over z minus 7")[$x = frac(y + 5, z − 7)$] for #math.equation(block: false, alt: "z .")[$z .$]

#solutionbox[
#math.equation(block: true, alt: "z equals the fraction y plus 5 plus 7 x over x")[$z = frac(y + 5 + 7 x, x)$]
]

#math.equation(block: false, alt: "P equals the fraction k over V")[$P = frac(k, V)$] for #math.equation(block: false, alt: "V .")[$V .$]


==== Solve Applications with Rational Equations

#strong[Solve Proportions]

In the following exercises, solve.

#math.equation(block: true, alt: "the fraction x over 4 equals the fraction 3 over 5")[$frac(x, 4) = frac(3, 5)$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 12 over 5")[$x = frac(12, 5)$]
]

#math.equation(block: true, alt: "the fraction 3 over y equals the fraction 9 over 5")[$frac(3, y) = frac(9, 5)$]

#math.equation(block: true, alt: "the fraction s over s plus 20 equals the fraction 3 over 7")[$frac(s, s + 20) = frac(3, 7)$]

#solutionbox[
#math.equation(block: true, alt: "s equals 15")[$s = 15$]
]

#math.equation(block: true, alt: "the fraction t minus 3 over 5 equals the fraction t plus 2 over 9")[$frac(t − 3, 5) = frac(t + 2, 9)$]

#strong[Solve Using Proportions]

In the following exercises, solve.

Rachael had a 21-ounce strawberry shake that has 739 calories. How many calories are there in a 32-ounce shake?

#solutionbox[
#math.equation(block: false, alt: "1126")[$1126$] calories
]

Leo went to Mexico over Christmas break and changed \$525 dollars into Mexican pesos. At that time, the exchange rate had \$1 US is equal to 16.25 Mexican pesos. How many Mexican pesos did he get for his trip?

#strong[Solve Similar Figure Applications]

In the following exercises, solve.

#math.equation(block: false, alt: "Δ A B C")[$"Δ" A B C$] is similar to #math.equation(block: false, alt: "Δ X Y Z .")[$"Δ" X Y Z .$] The lengths of two sides of each triangle are given in the figure. Find the lengths of the third sides. #linebreak() 
#figure(figph[The first figure is triangle A B C with side A B 8 units long, side B C 7 units long, and side A C b units long. The second figure is triangle X Y Z with side X Y 2 and two-thirds units long, side Y Z x units long, and side X Z 3 units long.], alt: "The first figure is triangle A B C with side A B 8 units long, side B C 7 units long, and side A C b units long. The second figure is triangle X Y Z with side X Y 2 and two-thirds units long, side Y Z x units long, and side X Z 3 units long.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "b equals 9 ; x equals 2 the fraction 1 over 3")[$b = 9 ; x = 2 frac(1, 3)$]
]

On a map of Europe, Paris, Rome, and Vienna form a triangle whose sides are shown in the figure below. If the actual distance from Rome to Vienna is 700 miles, find the distance from #linebreak() ⓐ Paris to Rome #linebreak() ⓑ Paris to Vienna #linebreak() 
#figure(figph[The figure is a triangle formed by Paris, Vienna, and Rome. The distance between Paris and Vienna is 7.7 centimeters. The distance between Vienna and Rome is 7 centimeters. The distance between Rome and Paris is 8.9 centimeters.], alt: "The figure is a triangle formed by Paris, Vienna, and Rome. The distance between Paris and Vienna is 7.7 centimeters. The distance between Vienna and Rome is 7 centimeters. The distance between Rome and Paris is 8.9 centimeters.", caption: none)

Francesca is 5.75 feet tall. Late one afternoon, her shadow was 8 feet long. At the same time, the shadow of a nearby tree was 32 feet long. Find the height of the tree.

#solutionbox[
23 feet
]

The height of a lighthouse in Pensacola, Florida is 150 feet. Standing next to the statue, 5.5-foot-tall Natasha cast a 1.1-foot shadow. How long would the shadow of the lighthouse be?

#strong[Solve Uniform Motion Applications]

In the following exercises, solve.

When making the 5-hour drive home from visiting her parents, Lolo ran into bad weather. She was able to drive 176 miles while the weather was good, but then driving 10 mph slower, went 81 miles when it turned bad. How fast did she drive when the weather was bad?

#solutionbox[
#math.equation(block: false, alt: "45")[$45$] mph
]

Mark is riding on a plane that can fly 490 miles with a headwind of 20 mph in the same time that it can fly 350 miles against a tailwind of 20 mph. What is the speed of the plane?

Josue can ride his bicycle 8 mph faster than Arjun can ride his bike. It takes Arjun 3 hours longer than Josue to ride 48 miles. How fast can Josue ride his bike?

#solutionbox[
#math.equation(block: false, alt: "16")[$16$] mph
]

Curtis was training for a triathlon. He ran 8 kilometers and biked 32 kilometers in a total of 3 hours. His running speed was 8 kilometers per hour less than his biking speed. What was his running speed?

#strong[Solve Work Applications]

In the following exercises, solve.

Brandy can frame a room in 1 hour, while Jake takes 4 hours. How long could they frame a room working together?

#solutionbox[
#math.equation(block: false, alt: "48")[$48$] minutes
]

Prem takes 3 hours to mow the lawn while her cousin, Barb, takes 2 hours. How long will it take them working together?

Jeffrey can paint a house in 6 days, but if he gets a helper he can do it in 4 days. How long would it take the helper to paint the house alone?

#solutionbox[
#math.equation(block: false, alt: "12")[$12$] days
]

Marta and Deb work together writing a book that takes them 90 days. If Marta worked alone it would take her 120 days. How long would it take Deb to write the book alone?

#strong[Solve Direct Variation Problems]

In the following exercises, solve.

If #math.equation(block: false, alt: "y")[$y$] varies directly as #math.equation(block: false, alt: "x")[$x$] when #math.equation(block: false, alt: "y equals 9")[$y = 9$] and #math.equation(block: false, alt: "x equals 3 ,")[$x = 3 ,$] find #math.equation(block: false, alt: "x")[$x$] when #math.equation(block: false, alt: "y equals 21 .")[$y = 21 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 7")[$x = 7$]
]

If #math.equation(block: false, alt: "y")[$y$] varies inversely as #math.equation(block: false, alt: "x")[$x$] when #math.equation(block: false, alt: "y equals 20")[$y = 20$] and #math.equation(block: false, alt: "x equals 2 ,")[$x = 2 ,$] find #math.equation(block: false, alt: "y")[$y$] when #math.equation(block: false, alt: "x equals 4 .")[$x = 4 .$]

Vanessa is traveling to see her fiancé. The distance, #math.equation(block: false, alt: "d ,")[$d ,$] varies directly with the speed, #math.equation(block: false, alt: "v ,")[$v ,$] she drives. If she travels 258 miles driving 60 mph, how far would she travel going 70 mph?

#solutionbox[
#math.equation(block: false, alt: "301")[$301$] mph
]

If the cost of a pizza varies directly with its diameter, and if an 8” diameter pizza costs \$12, how much would a 6” diameter pizza cost?

The distance to stop a car varies directly with the square of its speed. It takes 200 feet to stop a car going 50 mph. How many feet would it take to stop a car going 60 mph?

#solutionbox[
#math.equation(block: false, alt: "288")[$288$] feet
]

#strong[Solve Inverse Variation Problems]

In the following exercises, solve.

If #math.equation(block: false, alt: "m")[$m$] varies inversely with the square of #math.equation(block: false, alt: "n ,")[$n ,$] when #math.equation(block: false, alt: "m equals 4")[$m = 4$] and #math.equation(block: false, alt: "n equals 6")[$n = 6$] find #math.equation(block: false, alt: "m")[$m$] when #math.equation(block: false, alt: "n equals 2 .")[$n = 2 .$]

The number of tickets for a music fundraiser varies inversely with the price of the tickets. If Madelyn has just enough money to purchase 12 tickets for \$6 each, how many tickets can Madelyn afford to buy if the price increased to \$8?

#solutionbox[
#math.equation(block: false, alt: "9")[$9$] tickets
]

On a string instrument, the length of a string varies inversely with the frequency of its vibrations. If an 11-inch string on a violin has a frequency of 360 cycles per second, what frequency does a 12-inch string have?


==== Solve Rational Inequalities

#strong[Solve Rational Inequalities]

In the following exercises, solve each rational inequality and write the solution in interval notation.

#math.equation(block: true, alt: "the fraction x minus 3 over x plus 4 less than or equal to 0")[$frac(x − 3, x + 4) ≤ 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −4 , 3 ]")[$( −4 , 3 ]$]
]

#math.equation(block: true, alt: "the fraction 5 x over x minus 2 greater than 1")[$frac(5 x, x − 2) > 1$]

#math.equation(block: true, alt: "the fraction 3 x minus 2 over x minus 4 less than or equal to 2")[$frac(3 x − 2, x − 4) ≤ 2$]

#solutionbox[
#math.equation(block: true, alt: "[ −6 , 4 close parenthesis")[$[ −6 , 4 )$]
]

#math.equation(block: true, alt: "the fraction 1 over x squared minus 4 x minus 12 less than 0")[$frac(1, x^(2) − 4 x − 12) < 0$]

#math.equation(block: true, alt: "the fraction 1 over 2 minus the fraction 4 over x squared greater than or equal to the fraction 1 over x")[$frac(1, 2) − frac(4, x^(2)) ≥ frac(1, x)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −2 ] union [ 4 , ∞ close parenthesis")[$( "−" ∞ , −2 ] ∪ [ 4 , ∞ )$]
]

#math.equation(block: true, alt: "the fraction 4 over x minus 2 less than the fraction 3 over x plus 1")[$frac(4, x − 2) < frac(3, x + 1)$]

#strong[Solve an Inequality with Rational Functions]

In the following exercises, solve each rational function inequality and write the solution in interval notation

Given the function, #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x minus 5 over x minus 2 ,")[$R ( x ) = frac(x − 5, x − 2) ,$] find the values of #math.equation(block: false, alt: "x")[$x$] that make the function greater than or equal to 0.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , 2 close parenthesis union [ 5 , ∞ close parenthesis")[$( "−" ∞ , 2 ) ∪ [ 5 , ∞ )$]
]

Given the function, #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction x plus 1 over x plus 3 ,")[$R ( x ) = frac(x + 1, x + 3) ,$] find the values of #math.equation(block: false, alt: "x")[$x$] that make the function less than or equal to 0.

The function #linebreak() #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 150 x plus 100,000")[$C ( x ) = 150 x + 100,000$] represents the cost to produce #math.equation(block: false, alt: "x ,")[$x ,$] number of items. Find ⓐ the average cost function, #math.equation(block: false, alt: "c open parenthesis x close parenthesis")[$c ( x )$] ⓑ how many items should be produced so that the average cost is less than \$160.

#solutionbox[
ⓐ #math.equation(block: false, alt: "c open parenthesis x close parenthesis equals the fraction 150 x plus 100000 over x")[$c ( x ) = frac(150 x + 100000, x)$] #linebreak() ⓑ More than 10,000 items must be produced to keep the average cost below #math.equation(block: false, alt: "$ 160")[$"$" 160$] per item.
]

Tillman is starting his own business by selling tacos at the beach. Accounting for the cost of his food truck and ingredients for the tacos, the function #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 2 x plus 6,000")[$C ( x ) = 2 x + 6,000$] represents the cost for Tillman to produce #math.equation(block: false, alt: "x ,")[$x ,$] tacos. Find ⓐ the average cost function, #math.equation(block: false, alt: "c open parenthesis x close parenthesis")[$c ( x )$] for Tillman’s Tacos ⓑ how many tacos should Tillman produce so that the average cost is less than \$4.


=== Practice Test

In the following exercises, simplify.

#math.equation(block: true, alt: "the fraction 4 a squared b over 12 a b squared")[$frac(4 a^(2) b, 12 a b^(2))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction a over 3 b")[$frac(a, 3 b)$]
]

#math.equation(block: true, alt: "the fraction 6 x minus 18 over x squared minus 9")[$frac(6 x − 18, x^(2) − 9)$]

In the following exercises, perform the indicated operation and simplify.

#math.equation(block: true, alt: "the fraction 4 x over x plus 2 times the fraction x squared plus 5 x plus 6 over 12 x squared")[$frac(4 x, x + 2) · frac(x^(2) + 5 x + 6, 12 x^(2))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction x plus 3 over 3 x")[$frac(x + 3, 3 x)$]
]

#math.equation(block: true, alt: "the fraction 2 y squared over y squared minus 1 divided by the fraction y cubed minus y squared plus y over y cubed plus 1")[$frac(2 y^(2), y^(2) − 1) ÷ frac(y^(3) − y^(2) + y, y^(3) + 1)$]

#math.equation(block: true, alt: "the fraction 6 x squared minus x plus 20 over x squared minus 81 minus the fraction 5 x squared plus 11 x minus 7 over x squared minus 81")[$frac(6 x^(2) − x + 20, x^(2) − 81) − frac(5 x^(2) + 11 x − 7, x^(2) − 81)$]

#solutionbox[
#math.equation(block: true, alt: "the fraction x minus 3 over x plus 9")[$frac(x − 3, x + 9)$]
]

#math.equation(block: true, alt: "the fraction −3 a over 3 a minus 3 plus the fraction 5 a over a squared plus 3 a minus 4")[$frac(−3 a, 3 a − 3) + frac(5 a, a^(2) + 3 a − 4)$]

#math.equation(block: true, alt: "the fraction 2 n squared plus 8 n minus 1 over n squared minus 1 minus the fraction n squared minus 7 n minus 1 over 1 minus n squared")[$frac(2 n^(2) + 8 n − 1, n^(2) − 1) − frac(n^(2) − 7 n − 1, 1 − n^(2))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 3 n minus 2 over n minus 1")[$frac(3 n − 2, n − 1)$]
]

#math.equation(block: true, alt: "the fraction 10 x squared plus 16 x minus 7 over 8 x minus 3 plus the fraction 2 x squared plus 3 x minus 1 over 3 minus 8 x")[$frac(10 x^(2) + 16 x − 7, 8 x − 3) + frac(2 x^(2) + 3 x − 1, 3 − 8 x)$]

#math.equation(block: true, alt: "the fraction the fraction 1 over m minus the fraction 1 over n over the fraction 1 over n plus the fraction 1 over m")[$frac(frac(1, m) − frac(1, n), frac(1, n) + frac(1, m))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction n minus m over m plus n")[$frac(n − m, m + n)$]
]

In the following exercises, solve each equation.

#math.equation(block: true, alt: "the fraction 1 over x plus the fraction 3 over 4 equals the fraction 5 over 8")[$frac(1, x) + frac(3, 4) = frac(5, 8)$]

#math.equation(block: true, alt: "the fraction 1 over z minus 5 plus the fraction 1 over z plus 5 equals the fraction 1 over z squared minus 25")[$frac(1, z − 5) + frac(1, z + 5) = frac(1, z^(2) − 25)$]

#solutionbox[
#math.equation(block: true, alt: "z equals the fraction 1 over 2")[$z = frac(1, 2)$]
]

#math.equation(block: true, alt: "the fraction z over 2 z plus 8 minus the fraction 3 over 4 z minus 8 equals the fraction 3 z squared minus 16 z minus 16 over 8 z squared plus 16 z minus 64")[$frac(z, 2 z + 8) − frac(3, 4 z − 8) = frac(3 z^(2) − 16 z − 16, 8 z^(2) + 16 z − 64)$]

In the following exercises, solve each rational inequality and write the solution in interval notation.

#math.equation(block: true, alt: "the fraction 6 x over x minus 6 less than or equal to 2")[$frac(6 x, x − 6) ≤ 2$]

#solutionbox[
#math.equation(block: true, alt: "[ −3 , 6 close parenthesis")[$[ −3 , 6 )$]
]

#math.equation(block: true, alt: "the fraction 2 x plus 3 over x minus 6 greater than 1")[$frac(2 x + 3, x − 6) > 1$]

#math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 12 over x squared greater than or equal to the fraction 5 over x")[$frac(1, 2) + frac(12, x^(2)) ≥ frac(5, x)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , 0 close parenthesis union open parenthesis 0 , 4 ] union [ 6 , ∞ close parenthesis")[$( "−" ∞ , 0 ) ∪ ( 0 , 4 ] ∪ [ 6 , ∞ )$]
]

In the following exercises, find #math.equation(block: false, alt: "R open parenthesis x close parenthesis")[$R ( x )$] given #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x minus 4 over x squared minus 3 x minus 10")[$f ( x ) = frac(x − 4, x^(2) − 3 x − 10)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x minus 5 over x squared minus 2 x minus 8 .")[$g ( x ) = frac(x − 5, x^(2) − 2 x − 8) .$]

#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals f open parenthesis x close parenthesis minus g open parenthesis x close parenthesis")[$R ( x ) = f ( x ) − g ( x )$]

#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals f open parenthesis x close parenthesis times g open parenthesis x close parenthesis")[$R ( x ) = f ( x ) · g ( x )$]

#solutionbox[
#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals the fraction 1 over open parenthesis x plus 2 close parenthesis open parenthesis x plus 2 close parenthesis")[$R ( x ) = frac(1, ( x + 2 ) ( x + 2 ))$]
]

#math.equation(block: true, alt: "R open parenthesis x close parenthesis equals f open parenthesis x close parenthesis divided by g open parenthesis x close parenthesis")[$R ( x ) = f ( x ) ÷ g ( x )$]

Given the function, #linebreak() #math.equation(block: false, alt: "R open parenthesis x close parenthesis equals the fraction 2 over 2 x squared plus x minus 15 ,")[$R ( x ) = frac(2, 2 x^(2) + x − 15) ,$] find the values of #math.equation(block: false, alt: "x")[$x$] that make the function less than or equal to 0.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis −3 , the fraction 5 over 2 close parenthesis")[$( −3 , #h(0.2em) frac(5, 2) )$]
]

In the following exercises, solve.

If #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$], and #math.equation(block: false, alt: "x equals 5")[$x = 5$] when #math.equation(block: false, alt: "y equals 30 ,")[$y = 30 ,$] find #math.equation(block: false, alt: "x")[$x$] when #math.equation(block: false, alt: "y equals 42 .")[$y = 42 .$]

If #math.equation(block: false, alt: "y")[$y$] varies inversely with the square of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "x equals 3")[$x = 3$] when #math.equation(block: false, alt: "y equals 9 ,")[$y = 9 ,$] find #math.equation(block: false, alt: "y")[$y$] when #math.equation(block: false, alt: "x equals 4 .")[$x = 4 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 81 over 16")[$y = frac(81, 16)$]
]

Matheus can ride his bike for 30 miles with the wind in the same amount of time that he can go 21 miles against the wind. If the wind’s speed is 6 mph, what is Matheus’ speed on his bike?

Oliver can split a truckload of logs in 8 hours, but working with his dad they can get it done in 3 hours. How long would it take Oliver’s dad working alone to split the logs?

#solutionbox[
Oliver’s dad would take #math.equation(block: false, alt: "4 the fraction 4 over 5")[$4 frac(4, 5)$] hours to split the logs himself.
]

The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of 29.5 liters with 2000 psi, what is the volume if the tank has a 14.7 psi rating? Round to the nearest whole number.

The cities of Dayton, Columbus, and Cincinnati form a triangle in southern Ohio. The diagram gives the map distances between these cities in inches.

#figure(figph[The figure is a triangle formed by Cincinnati, Dayton, and Columbus. The distance between Cincinnati and Dayton is 2.4 inches. The distance between Dayton and Columbus is 3.2 inches. The distance between Columbus and Cincinnati is 5.3 inches.], alt: "The figure is a triangle formed by Cincinnati, Dayton, and Columbus. The distance between Cincinnati and Dayton is 2.4 inches. The distance between Dayton and Columbus is 3.2 inches. The distance between Columbus and Cincinnati is 5.3 inches.", caption: none)

The actual distance from Dayton to Cincinnati is 48 miles. What is the actual distance between Dayton and Columbus?

#solutionbox[
The distance between Dayton and Columbus is 64 miles.
]