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== 7.4#h(0.6em)Solve Rational Equations

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

Solve: #math.equation(block: false, alt: "the fraction 1 over 6 x plus the fraction 1 over 2 equals the fraction 1 over 3 .")[$frac(1, 6) x + frac(1, 2) = frac(1, 3) .$] #linebreak() 
If you missed this problem, review .

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "n squared minus 5 n minus 36 equals 0 .")[$n^(2) − 5 n − 36 = 0 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "n equals 9 , n equals −4")[$n = 9 , #h(0.2em) n = −4$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "5 x plus 2 y equals 10")[$5 x + 2 y = 10$] for #math.equation(block: false, alt: "y .")[$y .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 10 minus 5 x over 2")[$y = frac(10 − 5 x, 2)$]
]
]

After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have #emph[simplified] many kinds of #emph[expressions] and #emph[solved] many kinds of #emph[equations]. We have simplified many rational expressions so far in this chapter. Now we will #emph[solve] a #strong[rational equation].

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

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

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

#math.equation(block: true, alt: "Rational Expression, Rational Equation; the fraction 1 over 8 x plus the fraction 1 over 2; the fraction y plus 6 over y squared minus 36; the fraction 1 over n minus 3 plus the fraction 1 over n plus 4, the fraction 1 over 8 x plus the fraction 1 over 2, equals, the fraction 1 over 4; the fraction y plus 6 over y squared minus 36, equals, y plus 1; the fraction 1 over n minus 3 plus the fraction 1 over n plus 4, equals, the fraction 15 over n squared plus n minus 12")[$"Rational Expression" &  &  & "Rational Equation" \ frac(1, 8) x + frac(1, 2) \ frac(y + 6, y^(2) − 36) \ frac(1, n − 3) + frac(1, n + 4) &  &  & frac(1, 8) x + frac(1, 2) & = & frac(1, 4) \ frac(y + 6, y^(2) − 36) & = & y + 1 \ frac(1, n − 3) + frac(1, n + 4) & = & frac(15, n^(2) + n − 12)$]
=== Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an #strong[extraneous solution to a rational equation].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Extraneous Solution to a Rational Equation]

An #strong[extraneous solution to a rational equation] is an algebraic solution that would cause any of the expressions in the original equation to be undefined.
]

We note any possible extraneous solutions, #emph[c], by writing #math.equation(block: false, alt: "x not equal to c")[$x ≠ c$] next to the equation.

#examplebox("Example 1")[How to Solve a Rational Equation][
Solve: #math.equation(block: false, alt: "the fraction 1 over x plus the fraction 1 over 3 equals the fraction 5 over 6 .")[$frac(1, x) + frac(1, 3) = frac(5, 6) .$]

#solutionbox[
#figure(figph[Step 1 is to find any value of the variable that makes the denominator of the zero. Remember that if x is equal to 0, then 1 divided by x is undefined. So the equation becomes the sum of 1 divided by x and one-third is equal to five-sixths, where x is not equal to 0.], alt: "Step 1 is to find any value of the variable that makes the denominator of the zero. Remember that if x is equal to 0, then 1 divided by x is undefined. So the equation becomes the sum of 1 divided by x and one-third is equal to five-sixths, where x is not equal to 0.", caption: none)

#figure(figph[Step 2 is to find the least common denominator of all the fractions in the problem, 1 divided by x, one-third, and five-sixths. The least common denominator is 6 x.], alt: "Step 2 is to find the least common denominator of all the fractions in the problem, 1 divided by x, one-third, and five-sixths. The least common denominator is 6 x.", caption: none)

#figure(figph[Step 3 is to clear the fractions in the equation by multiplying each side by the least common denominator. The result is 6 x times the sum of 1 divided by x and one-third is equal to 6 x times five-sixths. Simplify using the distributive property. The result is 6 x times the quantity1 divided by x plus 6 x times one-third is equal to 6 x times five-sixths, which simplifies to 6 plus 2 x is equal to 5 x. This simplifies to 6 is equal to 3 x.], alt: "Step 3 is to clear the fractions in the equation by multiplying each side by the least common denominator. The result is 6 x times the sum of 1 divided by x and one-third is equal to 6 x times five-sixths. Simplify using the distributive property. The result is 6 x times the quantity1 divided by x plus 6 x times one-third is equal to 6 x times five-sixths, which simplifies to 6 plus 2 x is equal to 5 x. This simplifies to 6 is equal to 3 x.", caption: none)

#figure(figph[Step 4 is to solve the equation that results. The result is 2 is equal to x], alt: "Step 4 is to solve the equation that results. The result is 2 is equal to x", caption: none)

#figure(figph[Step 5 is to check the solution. Remember that any solutions that makes the original expression undefined must be discarded. The solution is not 0. Substitute x is equal to 2 into the original equation, 1 divided by x plus one-third is equal to five-sixths. Is one-half plus one-third is equal to five-sixths a true equation? Is three-sixths plus two-sixth is equal to five-sixths a true equation? Three-sixths plus two-sixth is equal to five-sixths. Five-sixth is equal to five-sixth is a true equation. So, the solution is x is equal to 2], alt: "Step 5 is to check the solution. Remember that any solutions that makes the original expression undefined must be discarded. The solution is not 0. Substitute x is equal to 2 into the original equation, 1 divided by x plus one-third is equal to five-sixths. Is one-half plus one-third is equal to five-sixths a true equation? Is three-sixths plus two-sixth is equal to five-sixths a true equation? Three-sixths plus two-sixth is equal to five-sixths. Five-sixth is equal to five-sixth is a true equation. So, the solution is x is equal to 2", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 1 over y plus the fraction 2 over 3 equals the fraction 1 over 5 .")[$frac(1, y) + frac(2, 3) = frac(1, 5) .$]

#solutionbox[
#math.equation(block: true, alt: "y equals minus the fraction 15 over 7")[$y = − frac(15, 7)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 2 over 3 plus the fraction 1 over 5 equals the fraction 1 over x .")[$frac(2, 3) + frac(1, 5) = frac(1, x) .$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 15 over 13")[$x = frac(15, 13)$]
]
]

The steps of this method are shown.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve equations with rational expressions.]

+ Note any value of the variable that would make any denominator zero.
+ Find the least common denominator of #emph[all] denominators in the equation.
+ Clear the fractions by multiplying both sides of the equation by the LCD.
+ Solve the resulting equation.
+ Check:
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.
]

We always start by noting the values that would cause any denominators to be zero.

#examplebox("Example 2")[How to Solve a Rational Equation using the Zero Product Property][
Solve: #math.equation(block: false, alt: "1 minus the fraction 5 over y equals minus the fraction 6 over y squared .")[$1 − frac(5, y) = − frac(6, y^(2)) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed: 1 - 5/y = -6/y^2. This is a rational equation that can be transformed into a quadratic equation by multiplying all terms by y^2 and rearranging them.], alt: "A mathematical equation is displayed: 1 - 5/y = -6/y^2. This is a rational equation that can be transformed into a quadratic equation by multiplying all terms by y^2 and rearranging them.", caption: none)]),
  [Note any value of the variable that would make #linebreak() any denominator zero.], [#figure(figph[A mathematical equation is displayed: 1 - 5/y = -6/y^2, with the condition that y is not equal to 0. This is a rational equation that can be solved for y.], alt: "A mathematical equation is displayed: 1 - 5/y = -6/y^2, with the condition that y is not equal to 0. This is a rational equation that can be solved for y.", caption: none)],
  [Find the least common denominator of all denominators in #linebreak() the equation. The LCD is #emph[y]#super[2].], [],
  [Clear the fractions by multiplying both sides of #linebreak() the equation by the LCD.], [#figure(figph[A mathematical equation is displayed, showing y^2 multiplied by (1 - 5/y) on the left side, and y^2 multiplied by (-6/y^2) on the right side, both separated by an equals sign.], alt: "A mathematical equation is displayed, showing y^2 multiplied by (1 - 5/y) on the left side, and y^2 multiplied by (-6/y^2) on the right side, both separated by an equals sign.", caption: none)],
  [Distribute.], [#figure(figph[A mathematical equation is displayed on a white background, reading 'y^2 \* 1 - y^2(5/y) = y^2(-6/y^2)'. The term 'y^2' is highlighted in red in the first two instances.], alt: "A mathematical equation is displayed on a white background, reading 'y^2 * 1 - y^2(5/y) = y^2(-6/y^2)'. The term 'y^2' is highlighted in red in the first two instances.", caption: none)],
  [Multiply.], [#figure(figph[The image shows the quadratic equation y^2 - 5y = -6.], alt: "The image shows the quadratic equation y^2 - 5y = -6.", caption: none)],
  [Solve the resulting equation. First #linebreak() write the quadratic equation in standard form.], [#figure(figph[A mathematical equation is displayed, reading y squared minus 5y plus 6 equals 0.], alt: "A mathematical equation is displayed, reading y squared minus 5y plus 6 equals 0.", caption: none)],
  [Factor.], [#figure(figph[A mathematical equation is displayed on a white background: (y-2)(y-3)=0. This quadratic equation is in factored form, indicating its roots are y=2 and y=3.], alt: "A mathematical equation is displayed on a white background: (y-2)(y-3)=0. This quadratic equation is in factored form, indicating its roots are y=2 and y=3.", caption: none)],
  [Use the Zero Product Property.], [#figure(figph[A mathematical expression states 'y - 2 = 0 or y - 3 = 0' in black text against a white background.], alt: "A mathematical expression states 'y - 2 = 0 or y - 3 = 0' in black text against a white background.", caption: none)],
  [Solve.], [#figure(figph[The image shows the mathematical expression y = 2 or y = 3, indicating two possible values for the variable y.], alt: "The image shows the mathematical expression y = 2 or y = 3, indicating two possible values for the variable y.", caption: none)],
  [Check. #linebreak() We did not get 0 as an algebraic solution. #linebreak()  #linebreak()  #linebreak() #figure(figph[Two columns demonstrate checking solutions y=2 and y=3 for the equation 1 - 5/y = 6/y^2, showing step-by-step calculations that confirm both values are correct.], alt: "Two columns demonstrate checking solutions y=2 and y=3 for the equation 1 - 5/y = 6/y^2, showing step-by-step calculations that confirm both values are correct.", caption: none)], [],
  [], [The solution is #math.equation(block: false, alt: "y equals 2 ,")[$y = 2 ,$] #math.equation(block: false, alt: "y equals 3 .")[$y = 3 .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "1 minus the fraction 2 over x equals the fraction 15 over x squared .")[$1 − frac(2, x) = frac(15, x^(2)) .$]

#solutionbox[
#math.equation(block: true, alt: "x equals −3 , x equals 5")[$x = −3 , x = 5$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "1 minus the fraction 4 over y equals the fraction 12 over y squared .")[$1 − frac(4, y) = frac(12, y^(2)) .$]

#solutionbox[
#math.equation(block: true, alt: "y equals −2 , y equals 6")[$y = −2 , y = 6$]
]
]

In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.

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

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A rational equation displayed on a white background, reading '2/(x+2) + 4/(x-2) = (x-1)/(x^2-4)'.], alt: "A rational equation displayed on a white background, reading '2/(x+2) + 4/(x-2) = (x-1)/(x^2-4)'.", caption: none)]),
  [Note any value of the variable #linebreak() that would make any denominator #linebreak() zero.], [#figure(figph[A rational algebraic equation is shown: 2/(x+2) + 4/(x-2) = (x-1)/((x+2)(x-2)), with the conditions x not equal to -2 and x not equal to 2, ensuring denominators are non-zero.], alt: "A rational algebraic equation is shown: 2/(x+2) + 4/(x-2) = (x-1)/((x+2)(x-2)), with the conditions x not equal to -2 and x not equal to 2, ensuring denominators are non-zero.", caption: none)],
  [Find the least common #linebreak() denominator of all denominators #linebreak() in the equation. #linebreak() The LCD is #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x minus 2 close parenthesis .")[$( x + 2 ) ( x − 2 ) .$]], [],
  [Clear the fractions by multiplying #linebreak() both sides of the equation by the #linebreak() LCD.], [#figure(figph[An algebraic equation is shown, with the term (x+2)(x-2) highlighted in red and appearing on both sides of the equality, multiplying other rational expressions.], alt: "An algebraic equation is shown, with the term (x+2)(x-2) highlighted in red and appearing on both sides of the equality, multiplying other rational expressions.", caption: none)],
  [Distribute.], [#figure(figph[An algebraic equation with rational expressions. Each term is multiplied by the common factor (x+2)(x-2), aiming to clear the denominators before solving for x.], alt: "An algebraic equation with rational expressions. Each term is multiplied by the common factor (x+2)(x-2), aiming to clear the denominators before solving for x.", caption: none)],
  [Remove common factors.], [#figure(figph[A mathematical equation demonstrating the process of clearing denominators by multiplying each term by the least common denominator, (x+2)(x-2), leading to cancellation of terms and simplification.], alt: "A mathematical equation demonstrating the process of clearing denominators by multiplying each term by the least common denominator, (x+2)(x-2), leading to cancellation of terms and simplification.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed against a white background: 2(x-2) + 4(x+2) = x-1. The equation involves a single variable 'x' with parentheses, multiplication, addition, and subtraction.], alt: "A mathematical equation is displayed against a white background: 2(x-2) + 4(x+2) = x-1. The equation involves a single variable 'x' with parentheses, multiplication, addition, and subtraction.", caption: none)],
  [Distribute.], [#figure(figph[A mathematical equation is displayed on a white background, which reads '2x - 4 + 4x + 8 = x - 1'.], alt: "A mathematical equation is displayed on a white background, which reads '2x - 4 + 4x + 8 = x - 1'.", caption: none)],
  [Solve.], [#figure(figph[A mathematical equation is displayed, showing '6x + 4 = x - 1' in black text against a plain white background, ready for solving.], alt: "A mathematical equation is displayed, showing '6x + 4 = x - 1' in black text against a plain white background, ready for solving.", caption: none)

 #linebreak() #figure(figph[A mathematical equation on a white background shows '5X = 5'.], alt: "A mathematical equation on a white background shows '5X = 5'.", caption: none)


 #linebreak() #figure(figph[An image displays the mathematical expression X = -1.], alt: "An image displays the mathematical expression X = -1.", caption: none)],
  [Check: #linebreak() We did not get 2 or −2 as algebraic solutions. #linebreak()  #linebreak() #figure(figph[Verifying the solution x = -1 for the rational equation. By substituting x = -1 into the original equation, both sides simplify to 2/3, confirming the validity of the solution.], alt: "Verifying the solution x = -1 for the rational equation. By substituting x = -1 into the original equation, both sides simplify to 2/3, confirming the validity of the solution.", caption: none)], [],
  [], [The solution is #math.equation(block: false, alt: "x equals −1 .")[$x = −1 .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 2 over x plus 1 plus the fraction 1 over x minus 1 equals the fraction 1 over x squared minus 1 .")[$frac(2, x + 1) + frac(1, x − 1) = frac(1, x^(2) − 1) .$]

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 5 over y plus 3 plus the fraction 2 over y minus 3 equals the fraction 5 over y squared minus 9 .")[$frac(5, y + 3) + frac(2, y − 3) = frac(5, y^(2) − 9) .$]

#solutionbox[
#math.equation(block: true, alt: "y equals 2")[$y = 2$]
]
]

In the next example, the first denominator is a #strong[trinomial]. Remember to factor it first to find the LCD.

#examplebox("Example 4")[][
Solve: #math.equation(block: false, alt: "the fraction m plus 11 over m squared minus 5 m plus 4 equals the fraction 5 over m minus 4 minus the fraction 3 over m minus 1 .")[$frac(m + 11, m^(2) − 5 m + 4) = frac(5, m − 4) − frac(3, m − 1) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[An algebraic equation showing (m+11) divided by (m^2-5m+4) equals 5 divided by (m-4) minus 3 divided by (m-1).], alt: "An algebraic equation showing (m+11) divided by (m^2-5m+4) equals 5 divided by (m-4) minus 3 divided by (m-1).", caption: none)]),
  [Note any value of the variable that #linebreak() would make any denominator zero. #linebreak() Use the factored form of the quadratic #linebreak() denominator.], [#figure(figph[A mathematical equation is displayed, showing a rational expression equal to the difference of two other rational expressions, with specified restrictions for the variable 'm'.], alt: "A mathematical equation is displayed, showing a rational expression equal to the difference of two other rational expressions, with specified restrictions for the variable 'm'.", caption: none)],
  [Find the least common denominator #linebreak() of all denominators in the equation. #linebreak() The LCD is #math.equation(block: false, alt: "open parenthesis m minus 4 close parenthesis open parenthesis m minus 1 close parenthesis .")[$( m − 4 ) ( m − 1 ) .$]], [],
  [Clear the fractions by #linebreak() multiplying both sides of the #linebreak() equation by the LCD.], [#figure(figph[An algebraic equation showing a step where both sides are multiplied by the common denominator (m-4)(m-1) to clear fractions, with the multiplying factor highlighted in red.], alt: "An algebraic equation showing a step where both sides are multiplied by the common denominator (m-4)(m-1) to clear fractions, with the multiplying factor highlighted in red.", caption: none)],
  [Distribute.], [#figure(figph[The algebraic simplification of an equation by multiplying each term by the common denominator (m-4)(m-1) to eliminate fractions and solve for m.], alt: "The algebraic simplification of an equation by multiplying each term by the common denominator (m-4)(m-1) to eliminate fractions and solve for m.", caption: none)],
  [Remove common factors.], [#figure(figph[An algebraic equation is shown undergoing simplification, with terms (m-4) and (m-1) visibly crossed out. This step demonstrates the cancellation of common factors to clear denominators and simplify the rational expression.], alt: "An algebraic equation is shown undergoing simplification, with terms (m-4) and (m-1) visibly crossed out. This step demonstrates the cancellation of common factors to clear denominators and simplify the rational expression.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed on a white background: m + 11 = 5(m - 1) - 3(m - 4).], alt: "A mathematical equation is displayed on a white background: m + 11 = 5(m - 1) - 3(m - 4).", caption: none)],
  [Solve the resulting equation.], [#figure(figph[A mathematical equation is displayed, showing m + 11 = 5m - 5 - 3m + 12 on a white background.], alt: "A mathematical equation is displayed, showing m + 11 = 5m - 5 - 3m + 12 on a white background.", caption: none)

 #linebreak() #figure(figph[4 = m], alt: "4 = m", caption: none)],
  [Check. #linebreak() The only algebraic solution #linebreak() was 4, but we said that 4 would make #linebreak() a denominator equal to zero. The algebraic solution is an #linebreak() extraneous solution.], [],
  [], [There is no solution to this equation.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction x plus 13 over x squared minus 7 x plus 10 equals the fraction 6 over x minus 5 minus the fraction 4 over x minus 2 .")[$frac(x + 13, x^(2) − 7 x + 10) = frac(6, x − 5) − frac(4, x − 2) .$]

#solutionbox[
There is no solution.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction y minus 6 over y squared plus 3 y minus 4 equals the fraction 2 over y plus 4 plus the fraction 7 over y minus 1 .")[$frac(y − 6, y^(2) + 3 y − 4) = frac(2, y + 4) + frac(7, y − 1) .$]

#solutionbox[
There is no solution.
]
]

The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.

#examplebox("Example 5")[][
Solve: #math.equation(block: false, alt: "the fraction y over y plus 6 equals the fraction 72 over y squared minus 36 plus 4 .")[$frac(y, y + 6) = frac(72, y^(2) − 36) + 4 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed on a white background, featuring the variable 'y' in a rational expression: y/(y+6) = 72/(y^2-36) + 4.], alt: "A mathematical equation is displayed on a white background, featuring the variable 'y' in a rational expression: y/(y+6) = 72/(y^2-36) + 4.", caption: none)]),
  [Factor all the denominators, #linebreak() so we can note any value of #linebreak() the variable that would make #linebreak() any denominator zero.], [#figure(figph[A mathematical equation is displayed: y/(y+6) = 72/((y-6)(y+6)) + 4, with conditions y ≠ 6 and y ≠ -6. The equation involves algebraic fractions and the variable 'y'.], alt: "A mathematical equation is displayed: y/(y+6) = 72/((y-6)(y+6)) + 4, with conditions y ≠ 6 and y ≠ -6. The equation involves algebraic fractions and the variable 'y'.", caption: none)],
  [Find the least common denominator. #linebreak() The LCD is #math.equation(block: false, alt: "open parenthesis y minus 6 close parenthesis open parenthesis y plus 6 close parenthesis .")[$( y − 6 ) ( y + 6 ) .$]], [],
  [Clear the fractions.], [#figure(figph[A mathematical equation where both sides are multiplied by (y-6)(y+6) to simplify a rational expression, likely a step in solving for y.], alt: "A mathematical equation where both sides are multiplied by (y-6)(y+6) to simplify a rational expression, likely a step in solving for y.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is shown: (y-6) \* y = 72 + (y-6)(y+6) \* 4.], alt: "A mathematical equation is shown: (y-6) * y = 72 + (y-6)(y+6) * 4.", caption: none)],
  [Simplify.], [#figure(figph[A quadratic equation is displayed, showing y(y-6) = 72 + 4(y^2-36).], alt: "A quadratic equation is displayed, showing y(y-6) = 72 + 4(y^2-36).", caption: none)],
  [Solve the resulting equation.], [#figure(figph[A quadratic equation is displayed: y^2 - 6y = 72 + 4y^2 - 144. This equation involves the variable 'y' raised to the power of two, along with linear terms and constants, making it suitable for algebraic manipulation to find the values of 'y' that satisfy the equation.], alt: "A quadratic equation is displayed: y^2 - 6y = 72 + 4y^2 - 144. This equation involves the variable 'y' raised to the power of two, along with linear terms and constants, making it suitable for algebraic manipulation to find the values of 'y' that satisfy the equation.", caption: none)

 #linebreak() #figure(figph[A mathematical equation is displayed, showing '0 = 3y² + 6y - 72' in a black font against a white background.], alt: "A mathematical equation is displayed, showing '0 = 3y² + 6y - 72' in a black font against a white background.", caption: none)


 #linebreak() #figure(figph[A mathematical equation is displayed on a white background: 0 = 3(y^2 + 2y - 24).], alt: "A mathematical equation is displayed on a white background: 0 = 3(y^2 + 2y - 24).", caption: none)


 #linebreak() #figure(figph[A mathematical equation is displayed, reading 0 = 3(y+6)(y-4). This equation represents a quadratic function in factored form, where the roots of y are -6 and 4.], alt: "A mathematical equation is displayed, reading 0 = 3(y+6)(y-4). This equation represents a quadratic function in factored form, where the roots of y are -6 and 4.", caption: none)


 #linebreak() #figure(figph[The image displays the equations y=-6 and y=4 written vertically on a white background, suggesting two distinct values for the variable y.], alt: "The image displays the equations y=-6 and y=4 written vertically on a white background, suggesting two distinct values for the variable y.", caption: none)],
  [Check. #linebreak()  #linebreak() #figure(figph[Verification of a mathematical solution. The image demonstrates checking if y=4 is a valid solution to the equation y/(y+6) = 72/(y^2-36) + 4, after y=-6 was identified as extraneous. The steps confirm that y=4 is a correct solution.], alt: "Verification of a mathematical solution. The image demonstrates checking if y=4 is a valid solution to the equation y/(y+6) = 72/(y^2-36) + 4, after y=-6 was identified as extraneous. The steps confirm that y=4 is a correct solution.", caption: none)], [],
  [], [The solution is #math.equation(block: false, alt: "y equals 4 .")[$y = 4 .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction x over x plus 4 equals the fraction 32 over x squared minus 16 plus 5 .")[$frac(x, x + 4) = frac(32, x^(2) − 16) + 5 .$]

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction y over y plus 8 equals the fraction 128 over y squared minus 64 plus 9 .")[$frac(y, y + 8) = frac(128, y^(2) − 64) + 9 .$]

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

In some cases, all the algebraic solutions are extraneous.

#examplebox("Example 6")[][
Solve: #math.equation(block: false, alt: "the fraction x over 2 x minus 2 minus the fraction 2 over 3 x plus 3 equals the fraction 5 x squared minus 2 x plus 9 over 12 x squared minus 12 .")[$frac(x, 2 x − 2) − frac(2, 3 x + 3) = frac(5 x^(2) − 2 x + 9, 12 x^(2) − 12) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#figure(figph[An algebraic equation showing the subtraction of two rational expressions equal to a third rational expression. The equation is x/(2x-2) - 2/(3x+3) = (5x^2-2x+9)/(12x^2-12).], alt: "An algebraic equation showing the subtraction of two rational expressions equal to a third rational expression. The equation is x/(2x-2) - 2/(3x+3) = (5x^2-2x+9)/(12x^2-12).", caption: none)],
  [We will start by factoring all #linebreak() denominators, to make it easier #linebreak() to identify extraneous solutions and the LCD.], [#figure(figph[An algebraic equation showing the difference of two rational expressions equal to a third rational expression. The equation is x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)).], alt: "An algebraic equation showing the difference of two rational expressions equal to a third rational expression. The equation is x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)).", caption: none)],
  [Note any value of the variable #linebreak() that would make any denominator zero.], [#figure(figph[A complex algebraic equation is displayed, involving rational expressions with variable x in both numerators and denominators. It states that x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)), with restrictions x ≠ 1 and x ≠ -1.], alt: "A complex algebraic equation is displayed, involving rational expressions with variable x in both numerators and denominators. It states that x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)), with restrictions x ≠ 1 and x ≠ -1.", caption: none)],
  [Find the least common #linebreak() denominator. #linebreak() The LCD is #math.equation(block: false, alt: "12 open parenthesis x minus 1 close parenthesis open parenthesis x plus 1 close parenthesis .")[$12 ( x − 1 ) ( x + 1 ) .$]], [],
  [Clear the fractions.], [#figure(figph[An algebraic equation illustrating the step of multiplying both sides by the common factor 12(x-1)(x+1). The left side consists of a difference of two fractions, and the right side is a single fractional term.], alt: "An algebraic equation illustrating the step of multiplying both sides by the common factor 12(x-1)(x+1). The left side consists of a difference of two fractions, and the right side is a single fractional term.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed on a white background: 6(x+1)x - 4(x-1)2 = 5x^2 - 2x + 9. Some parts of the equation, specifically the variables 'x' within parentheses and the '1's, are highlighted in red.], alt: "A mathematical equation is displayed on a white background: 6(x+1)x - 4(x-1)2 = 5x^2 - 2x + 9. Some parts of the equation, specifically the variables 'x' within parentheses and the '1's, are highlighted in red.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed on a white background: 6x(x+1) - 4 \* 2(x-1) = 5x^2 - 2x + 9. The numbers and variables are in a dark gray font.], alt: "A mathematical equation is displayed on a white background: 6x(x+1) - 4 * 2(x-1) = 5x^2 - 2x + 9. The numbers and variables are in a dark gray font.", caption: none)],
  [Solve the resulting equation.], [#figure(figph[A mathematical equation shows 6x squared plus 6x minus 8x plus 8 equals 5x squared minus 2x plus 9, appearing on a white background.], alt: "A mathematical equation shows 6x squared plus 6x minus 8x plus 8 equals 5x squared minus 2x plus 9, appearing on a white background.", caption: none)

 #linebreak() #figure(figph[The mathematical equation x^2 - 1 = 0 is displayed on a white background, suggesting a problem or concept from algebra or pre-calculus.], alt: "The mathematical equation x^2 - 1 = 0 is displayed on a white background, suggesting a problem or concept from algebra or pre-calculus.", caption: none)


 #linebreak() #figure(figph[The image displays the algebraic equation (x-1)(x+1) = 0, representing a quadratic equation in factored form.], alt: "The image displays the algebraic equation (x-1)(x+1) = 0, representing a quadratic equation in factored form.", caption: none)


 #linebreak() #figure(figph[The mathematical expression 'x = 1 or x = -1' is displayed vertically on a white background.], alt: "The mathematical expression 'x = 1 or x = -1' is displayed vertically on a white background.", caption: none)],
  [Check. #linebreak()  #linebreak()  #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals −1")[$x = −1$] are extraneous solutions.], [],
  [], [The equation has no solution.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction y over 5 y minus 10 minus the fraction 5 over 3 y plus 6 equals the fraction 2 y squared minus 19 y plus 54 over 15 y squared minus 60 .")[$frac(y, 5 y − 10) − frac(5, 3 y + 6) = frac(2 y^(2) − 19 y + 54, 15 y^(2) − 60) .$]

#solutionbox[
There is no solution.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, 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 .")[$frac(z, 2 z + 8) − frac(3, 4 z − 8) = frac(3 z^(2) − 16 z − 16, 8 z^(2) + 16 z − 64) .$]

#solutionbox[
There is no solution.
]
]

#examplebox("Example 7")[][
Solve: #math.equation(block: false, alt: "the fraction 4 over 3 x squared minus 10 x plus 3 plus the fraction 3 over 3 x squared plus 2 x minus 1 equals the fraction 2 over x squared minus 2 x minus 3 .")[$frac(4, 3 x^(2) − 10 x + 3) + frac(3, 3 x^(2) + 2 x − 1) = frac(2, x^(2) − 2 x − 3) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#figure(figph[A complex algebraic equation is displayed, showing the sum of two fractions equaling a third. The equation is 4/(3x^2 - 10x + 3) + 3/(3x^2 + 2x - 1) = 2/(x^2 - 2x - 3).], alt: "A complex algebraic equation is displayed, showing the sum of two fractions equaling a third. The equation is 4/(3x^2 - 10x + 3) + 3/(3x^2 + 2x - 1) = 2/(x^2 - 2x - 3).", caption: none)],
  [Factor all the denominators, so we can note any value of the variable that would make any denominator #linebreak() zero.], [#figure(figph[An algebraic equation showing the sum of two fractions equaling a third fraction: 4 over (3x-1)(x-3) plus 3 over (3x-1)(x+1) equals 2 over (x-3)(x+1).], alt: "An algebraic equation showing the sum of two fractions equaling a third fraction: 4 over (3x-1)(x-3) plus 3 over (3x-1)(x+1) equals 2 over (x-3)(x+1).", caption: none)

 #linebreak() 
#math.equation(block: false, alt: "x not equal to − 1 , x not equal to the fraction 1 over 3 , x not equal to 3")[$x ≠ "−" 1 , x ≠ frac(1, 3) , x ≠ 3$]],
  [Find the least common denominator. The LCD is #math.equation(block: false, alt: "open parenthesis 3 x minus 1 close parenthesis open parenthesis x plus 1 close parenthesis open parenthesis x minus 3 close parenthesis .")[$( 3 x − 1 ) ( x + 1 ) ( x − 3 ) .$]], [],
  [Clear the fractions.], [],
  [#figure(figph[An algebraic equation showing the product of three binomials, (3x-1)(x+1)(x-3), multiplied by a sum of two fractions on the left, equating to the same product multiplied by a single fraction on the right.], alt: "An algebraic equation showing the product of three binomials, (3x-1)(x+1)(x-3), multiplied by a sum of two fractions on the left, equating to the same product multiplied by a single fraction on the right.", caption: none)], [],
  [Simplify.], [#figure(figph[A mathematical equation is displayed, showing '4(x+1) + 3(x-3) = 2(3x-1)'.], alt: "A mathematical equation is displayed, showing '4(x+1) + 3(x-3) = 2(3x-1)'.", caption: none)],
  [Distribute.], [#figure(figph[A mathematical equation is displayed on a white background, which reads '4x + 4 + 3x - 9 = 6x - 2'.], alt: "A mathematical equation is displayed on a white background, which reads '4x + 4 + 3x - 9 = 6x - 2'.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed on a white background, reading '7x - 5 = 6x - 2' in black text.], alt: "A mathematical equation is displayed on a white background, reading '7x - 5 = 6x - 2' in black text.", caption: none)],
  [], [#figure(figph[The equation 'x = 3' is displayed in the center of a plain white background.], alt: "The equation 'x = 3' is displayed in the center of a plain white background.", caption: none)],
  [The only algebraic solution was #math.equation(block: false, alt: "x equals 3 ,")[$x = 3 ,$] but we said that #math.equation(block: false, alt: "x equals 3")[$x = 3$] would make a denominator equal to zero. The algebraic solution is an extraneous solution.], [],
  [], [There is no solution to this equation.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 15 over x squared plus x minus 6 minus the fraction 3 over x minus 2 equals the fraction 2 over x plus 3 .")[$frac(15, x^(2) + x − 6) − frac(3, x − 2) = frac(2, x + 3) .$]

#solutionbox[
There is no solution.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 5 over x squared plus 2 x minus 3 minus the fraction 3 over x squared plus x minus 2 equals the fraction 1 over x squared plus 5 x plus 6 .")[$frac(5, x^(2) + 2 x − 3) − frac(3, x^(2) + x − 2) = frac(1, x^(2) + 5 x + 6) .$]

#solutionbox[
There is no solution.
]
]


=== Use Rational Functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

#examplebox("Example 8")[][
For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 x minus 6 over x squared minus 8 x plus 15 ,")[$f ( x ) = frac(2 x − 6, x^(2) − 8 x + 15) ,$] ⓐ find the domain of the function, ⓑ solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1 ,")[$f ( x ) = 1 ,$] and ⓒ find the points on the graph at this function value.

#solutionbox[
ⓐ The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared minus 8 x plus 15 equals 0")[$x^(2) − 8 x + 15 = 0$]]),
  [Factor the trinomial.], [#math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x minus 5 close parenthesis equals 0")[$( x − 3 ) ( x − 5 ) = 0$]],
  [Use the Zero Product Property.], [#math.equation(block: false, alt: "x minus 3 equals 0 x minus 5 equals 0")[$x − 3 = 0 #h(1em) x − 5 = 0$]],
  [Solve.], [#math.equation(block: false, alt: "x equals 3 x equals 5")[$x = 3 #h(1em) x = 5$]],
  [], [The domain is all real numbers except #math.equation(block: false, alt: "x not equal to 3 , x not equal to 5 .")[$x ≠ 3 , x ≠ 5 "."$]],
))

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The mathematical equation f(x) = 1 is displayed in a clean, sans-serif font on a white background.], alt: "The mathematical equation f(x) = 1 is displayed in a clean, sans-serif font on a white background.", caption: none)]),
  [Substitute in the rational expression.], [#figure(figph[A rational equation is displayed: (2x - 6) / (x^2 - 8x + 15) = 1. The equation involves a linear expression in the numerator and a quadratic expression in the denominator, set equal to 1.], alt: "A rational equation is displayed: (2x - 6) / (x^2 - 8x + 15) = 1. The equation involves a linear expression in the numerator and a quadratic expression in the denominator, set equal to 1.", caption: none)],
  [Factor the denominator.], [#figure(figph[The image displays a mathematical equation: a fraction with (2x - 6) in the numerator and the product of (x - 3) and (x - 5) in the denominator, set equal to 1. This is a rational equation to be solved for x.], alt: "The image displays a mathematical equation: a fraction with (2x - 6) in the numerator and the product of (x - 3) and (x - 5) in the denominator, set equal to 1. This is a rational equation to be solved for x.", caption: none)],
  [Multiply both sides by the LCD, #linebreak() #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x minus 5 close parenthesis .")[$( x − 3 ) ( x − 5 ) .$]], [#figure(figph[An algebraic step demonstrating the multiplication of both sides of an equation by the common denominator (x-3)(x-5) to eliminate the rational expression.], alt: "An algebraic step demonstrating the multiplication of both sides of an equation by the common denominator (x-3)(x-5) to eliminate the rational expression.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed, showing '2x - 6 = x^2 - 8x + 15', a quadratic equation that can be rearranged and solved for x.], alt: "A mathematical equation is displayed, showing '2x - 6 = x^2 - 8x + 15', a quadratic equation that can be rearranged and solved for x.", caption: none)],
  [Solve.], [#figure(figph[A mathematical equation is displayed against a white background. The equation reads '0 = x^2 - 10x + 21'.], alt: "A mathematical equation is displayed against a white background. The equation reads '0 = x^2 - 10x + 21'.", caption: none)],
  [Factor.], [#figure(figph[A mathematical equation is shown with a blurred background. The equation reads '0 = (x - 7)(x - 3)', representing a quadratic equation in factored form.], alt: "A mathematical equation is shown with a blurred background. The equation reads '0 = (x - 7)(x - 3)', representing a quadratic equation in factored form.", caption: none)],
  [Use the Zero Product Property.], [#figure(figph[Two mathematical expressions are displayed horizontally. The expression on the left reads 'x - 7 = 0' and the expression on the right reads 'x - 3 = 0'.], alt: "Two mathematical expressions are displayed horizontally. The expression on the left reads 'x - 7 = 0' and the expression on the right reads 'x - 3 = 0'.", caption: none)],
  [Solve.], [#figure(figph[Two mathematical expressions are displayed against a white background: 'x = 7' on the left and 'x = 3' on the right, both in dark grey font.], alt: "Two mathematical expressions are displayed against a white background: 'x = 7' on the left and 'x = 3' on the right, both in dark grey font.", caption: none)],
))

However, #math.equation(block: false, alt: "x equals 3")[$x = 3$] is outside the domain of this function, so we discard that root as extraneous.

ⓒ The value of the function is 1 when #math.equation(block: false, alt: "x equals 7 .")[$x = 7 .$] So the points on the graph of this function when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1")[$f ( x ) = 1$] is #math.equation(block: false, alt: "open parenthesis 7 , 1 close parenthesis close parenthesis")[$\( 7 , 1 \) \)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 8 minus x over x squared minus 7 x plus 12 ,")[$f ( x ) = frac(8 − x, x^(2) − 7 x + 12) ,$] ⓐ find the domain of the function ⓑ solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3")[$f ( x ) = 3$] ⓒ 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 3")[$x ≠ 3$] and #math.equation(block: false, alt: "x not equal to 4 .")[$x ≠ 4 .$] ⓑ #math.equation(block: false, alt: "x equals 2 , x equals the fraction 14 over 3")[$x = 2 , x = frac(14, 3)$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis 2 , 3 close parenthesis , open parenthesis the fraction 14 over 3 , 3 close parenthesis")[$( 2 , 3 ) , ( frac(14, 3) , 3 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x minus 1 over x squared minus 6 x plus 5 ,")[$f ( x ) = frac(x − 1, x^(2) − 6 x + 5) ,$] ⓐ find the domain of the function ⓑ solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4")[$f ( x ) = 4$] ⓒ 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 1")[$x ≠ 1$] and #math.equation(block: false, alt: "x not equal to 5 .")[$x ≠ 5 .$] ⓑ #math.equation(block: false, alt: "x equals the fraction 21 over 4")[$x = frac(21, 4)$] ⓒ #math.equation(block: false, alt: "open parenthesis the fraction 21 over 4 , 4 close parenthesis")[$( frac(21, 4) , 4 )$]
]
]


=== Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.

#math.equation(block: true, alt: "m, equals, the fraction y minus y sub 1 over x minus x sub 1; Multiply both sides of the equation by x minus x sub 1 ., m open parenthesis x minus x sub 1 close parenthesis, equals, open parenthesis the fraction y minus y sub 1 over x minus x sub 1 close parenthesis open parenthesis x minus x sub 1 close parenthesis; Simplify., m open parenthesis x minus x sub 1 close parenthesis, equals, y minus y sub 1; Rewrite the equation with the y terms on the left., y minus y sub 1, equals, m open parenthesis x minus x sub 1 close parenthesis")[$&  &  & m & = & frac(y − y_(1), x − x_(1)) \ "Multiply both sides of the equation by" #h(0.2em) x − x_(1) . &  &  & m ( x − x_(1) ) & = & ( frac(y − y_(1), x − x_(1)) ) ( x − x_(1) ) \ "Simplify." &  &  & m ( x − x_(1) ) & = & y − y_(1) \ "Rewrite the equation with the" #h(0.2em) y #h(0.2em) "terms on the left." &  &  & y − y_(1) & = & m ( x − x_(1) )$]

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through a point in Chapter 3. We will add one more step to solve for #emph[y].

#examplebox("Example 9")[][
Solve:#math.equation(block: false, alt: "m equals the fraction y minus 2 over x minus 3")[$m = frac(y − 2, x − 3)$] for #math.equation(block: false, alt: "y .")[$y .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation shows the slope 'm' defined as a fraction: (y - 2) over (x - 3). The variable 'y' in the numerator is highlighted in red.], alt: "A mathematical equation shows the slope 'm' defined as a fraction: (y - 2) over (x - 3). The variable 'y' in the numerator is highlighted in red.", caption: none)]),
  [Note any value of the variable that would #linebreak() make any denominator zero.], [#figure(figph[A mathematical equation for slope is shown as m = (y - 2) / (x - 3), with the condition x ≠ 3 to prevent division by zero. The variable 'y' is highlighted in red.], alt: "A mathematical equation for slope is shown as m = (y - 2) / (x - 3), with the condition x ≠ 3 to prevent division by zero. The variable 'y' is highlighted in red.", caption: none)],
  [Clear the fractions by multiplying both sides of #linebreak() the equation by the LCD, #math.equation(block: false, alt: "x minus 3 .")[$x − 3 .$]], [#figure(figph[A mathematical equation is displayed, showing (x - 3)m = (x - 3) multiplied by the fraction (y - 2) over (x - 3). The variable 'y' in the numerator is highlighted in red.], alt: "A mathematical equation is displayed, showing (x - 3)m = (x - 3) multiplied by the fraction (y - 2) over (x - 3). The variable 'y' in the numerator is highlighted in red.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed with the text 'xm - 3m = y - 2' in black and red lettering on a white background. The variable 'y' is highlighted in red.], alt: "A mathematical equation is displayed with the text 'xm - 3m = y - 2' in black and red lettering on a white background. The variable 'y' is highlighted in red.", caption: none)],
  [Isolate the term with #emph[y].], [#figure(figph[A mathematical equation is displayed, reading 'xm - 3m + 2 = y' with the 'y' highlighted in red. The equation appears in a dark gray font against a plain white background.], alt: "A mathematical equation is displayed, reading 'xm - 3m + 2 = y' with the 'y' highlighted in red. The equation appears in a dark gray font against a plain white background.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "m equals the fraction y minus 5 over x minus 4")[$m = frac(y − 5, x − 4)$]for #math.equation(block: false, alt: "y .")[$y .$]

#solutionbox[
#math.equation(block: true, alt: "y equals m x minus 4 m plus 5")[$y = m x − 4 m + 5$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "m equals the fraction y minus 1 over x plus 5")[$m = frac(y − 1, x + 5)$] for #math.equation(block: false, alt: "y .")[$y .$]

#solutionbox[
#math.equation(block: true, alt: "y equals m x plus 5 m plus 1")[$y = m x + 5 m + 1$]
]
]

Remember to multiply both sides by the LCD in the next example.

#examplebox("Example 10")[][
Solve: #math.equation(block: false, alt: "the fraction 1 over c plus the fraction 1 over m equals 1")[$frac(1, c) + frac(1, m) = 1$] for #emph[c].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#figure(figph[A mathematical equation is displayed, showing '1/c + 1/m = 1 for c' in black text on a white background.], alt: "A mathematical equation is displayed, showing '1/c + 1/m = 1 for c' in black text on a white background.", caption: none)],
  [Note any value of the variable that would make #linebreak() any denominator zero.], [#figure(figph[A mathematical equation displays '1/c + 1/m = 1, c not equal 0, m not equal 0'.], alt: "A mathematical equation displays '1/c + 1/m = 1, c not equal 0, m not equal 0'.", caption: none)],
  [Clear the fractions by multiplying both sides of #linebreak() the equations by the LCD, #emph[cm].], [#figure(figph[Mathematical equation: cm(1/c + 1/m) = cm(1). The expression shows a step in algebraic simplification, where the terms inside the parentheses simplify to 1.], alt: "Mathematical equation: cm(1/c + 1/m) = cm(1). The expression shows a step in algebraic simplification, where the terms inside the parentheses simplify to 1.", caption: none)],
  [Distribute.], [#figure(figph[A mathematical equation is displayed against a white background, reading 'cm(1/c) + cm(1/m) = cm(1)', with 'cm' in red and the rest in black.], alt: "A mathematical equation is displayed against a white background, reading 'cm(1/c) + cm(1/m) = cm(1)', with 'cm' in red and the rest in black.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation on a white background reads 'm + c = cm'.], alt: "A mathematical equation on a white background reads 'm + c = cm'.", caption: none)],
  [Collect the terms with #emph[c] to the right.], [#figure(figph[A mathematical equation shows 'm = 500 - (', suggesting an incomplete expression or problem where 'm' is being defined in relation to '500' minus some value within parentheses.], alt: "A mathematical equation shows 'm = 500 - (', suggesting an incomplete expression or problem where 'm' is being defined in relation to '500' minus some value within parentheses.", caption: none)],
  [Factor the expression on the right.], [#figure(figph[The image displays a mathematical equation, m = c(m - 1), on a white background. This equation relates the variables 'm' and 'c', suggesting a problem or concept in algebra or mathematics.], alt: "The image displays a mathematical equation, m = c(m - 1), on a white background. This equation relates the variables 'm' and 'c', suggesting a problem or concept in algebra or mathematics.", caption: none)],
  [To isolate #emph[c], divide both sides by #math.equation(block: false, alt: "m minus 1 .")[$m − 1 .$]], [#figure(figph[The equation m/(m-1) = c(m-1)/(m-1) is displayed, with the common denominator 'm-1' highlighted in red.], alt: "The equation m/(m-1) = c(m-1)/(m-1) is displayed, with the common denominator 'm-1' highlighted in red.", caption: none)],
  [Simplify by removing common factors.], [#figure(figph[A mathematical equation is displayed, showing a fraction m divided by (m - 1), which is set equal to c. The equation reads: m/(m-1) = c.], alt: "A mathematical equation is displayed, showing a fraction m divided by (m - 1), which is set equal to c. The equation reads: m/(m-1) = c.", caption: none)],
  [Notice that even though we excluded #math.equation(block: false, alt: "c equals 0 , m equals 0")[$c = 0 , m = 0$] from the original equation, we must also now state that #math.equation(block: false, alt: "m not equal to 1 .")[$m ≠ 1 .$]], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 1 over a plus the fraction 1 over b equals c")[$frac(1, a) + frac(1, b) = c$] for #emph[a].

#solutionbox[
#math.equation(block: true, alt: "a equals the fraction b over c b minus 1")[$a = frac(b, c b − 1)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 2 over x plus the fraction 1 over 3 equals the fraction 1 over y")[$frac(2, x) + frac(1, 3) = frac(1, y)$] for #emph[y].

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access this online resource for additional instruction and practice with equations with rational expressions.

- #link("https://openstax.org/l/37EqRatExp")[Equations with Rational Expressions]
]


=== Key Concepts

- #strong[How to solve equations with rational expressions.]
+ Note any value of the variable that would make any denominator zero.
+ Find the least common denominator of all denominators in the equation.
+ Clear the fractions by multiplying both sides of the equation by the LCD.
+ Solve the resulting equation.
+ Check:
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.


==== Practice Makes Perfect

#strong[Solve Rational Equations]

In the following exercises, solve each rational equation.

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

#solutionbox[
#math.equation(block: true, alt: "a equals 10")[$a = 10$]
]

#math.equation(block: true, alt: "the fraction 6 over 3 minus the fraction 2 over d equals the fraction 4 over 9")[$frac(6, 3) − frac(2, d) = frac(4, 9)$]

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

#solutionbox[
#math.equation(block: true, alt: "v equals the fraction 40 over 21")[$v = frac(40, 21)$]
]

#math.equation(block: true, alt: "the fraction 3 over 8 plus the fraction 2 over y equals the fraction 1 over 4")[$frac(3, 8) + frac(2, y) = frac(1, 4)$]

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

#solutionbox[
#math.equation(block: true, alt: "m equals −2 , m equals 4")[$m = −2 , m = 4$]
]

#math.equation(block: true, alt: "1 plus the fraction 4 over n equals the fraction 21 over n squared")[$1 + frac(4, n) = frac(21, n^(2))$]

#math.equation(block: true, alt: "1 plus the fraction 9 over p equals the fraction −20 over p squared")[$1 + frac(9, p) = frac(−20, p^(2))$]

#solutionbox[
#math.equation(block: true, alt: "p equals −5 , p equals −4")[$p = −5 , p = −4$]
]

#math.equation(block: true, alt: "1 minus the fraction 7 over q equals the fraction −6 over q squared")[$1 − frac(7, q) = frac(−6, q^(2))$]

#math.equation(block: true, alt: "the fraction 5 over 3 v minus 2 equals the fraction 7 over 4 v")[$frac(5, 3 v − 2) = frac(7, 4 v)$]

#solutionbox[
#math.equation(block: true, alt: "v equals 14")[$v = 14$]
]

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

#math.equation(block: true, alt: "the fraction 3 over x plus 4 plus the fraction 7 over x minus 4 equals the fraction 8 over x squared minus 16")[$frac(3, x + 4) + frac(7, x − 4) = frac(8, x^(2) − 16)$]

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

#math.equation(block: true, alt: "the fraction 5 over y minus 9 plus the fraction 1 over y plus 9 equals the fraction 18 over y squared minus 81")[$frac(5, y − 9) + frac(1, y + 9) = frac(18, y^(2) − 81)$]

#math.equation(block: true, alt: "the fraction 8 over z minus 10 minus the fraction 7 over z plus 10 equals the fraction 5 over z squared minus 100")[$frac(8, z − 10) − frac(7, z + 10) = frac(5, z^(2) − 100)$]

#solutionbox[
#math.equation(block: true, alt: "z equals −145")[$z = −145$]
]

#math.equation(block: true, alt: "the fraction 9 over a plus 11 minus the fraction 6 over a minus 11 equals the fraction 6 over a squared minus 121")[$frac(9, a + 11) − frac(6, a − 11) = frac(6, a^(2) − 121)$]

#math.equation(block: true, alt: "the fraction −10 over q minus 2 minus the fraction 7 over q plus 4 equals 1")[$frac(−10, q − 2) − frac(7, q + 4) = 1$]

#solutionbox[
#math.equation(block: true, alt: "q equals −18 , q equals −1")[$q = −18 , q = −1$]
]

#math.equation(block: true, alt: "the fraction 2 over s plus 7 minus the fraction 3 over s minus 3 equals 1")[$frac(2, s + 7) − frac(3, s − 3) = 1$]

#math.equation(block: true, alt: "the fraction v minus 10 over v squared minus 5 v plus 4 equals the fraction 3 over v minus 1 minus the fraction 6 over v minus 4")[$frac(v − 10, v^(2) − 5 v + 4) = frac(3, v − 1) − frac(6, v − 4)$]

#solutionbox[
#math.equation(block: true, alt: "no solution")[$"no solution"$]
]

#math.equation(block: true, alt: "the fraction w plus 8 over w squared minus 11 w plus 28 equals the fraction 5 over w minus 7 plus the fraction 2 over w minus 4")[$frac(w + 8, w^(2) − 11 w + 28) = frac(5, w − 7) + frac(2, w − 4)$]

#math.equation(block: true, alt: "the fraction x minus 10 over x squared plus 8 x plus 12 equals the fraction 3 over x plus 2 plus the fraction 4 over x plus 6")[$frac(x − 10, x^(2) + 8 x + 12) = frac(3, x + 2) + frac(4, x + 6)$]

#solutionbox[
#math.equation(block: true, alt: "no solution")[$"no solution"$]
]

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

#math.equation(block: true, alt: "the fraction b plus 3 over 3 b plus the fraction b over 24 equals the fraction 1 over b")[$frac(b + 3, 3 b) + frac(b, 24) = frac(1, b)$]

#solutionbox[
#math.equation(block: true, alt: "b equals −8")[$b = −8$]
]

#math.equation(block: true, alt: "the fraction c plus 3 over 12 c plus the fraction c over 36 equals the fraction 1 over 4 c")[$frac(c + 3, 12 c) + frac(c, 36) = frac(1, 4 c)$]

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

#solutionbox[
#math.equation(block: true, alt: "d equals 2")[$d = 2$]
]

#math.equation(block: true, alt: "the fraction m over m plus 5 equals the fraction 50 over m squared minus 25 plus 6")[$frac(m, m + 5) = frac(50, m^(2) − 25) + 6$]

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

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

#math.equation(block: true, alt: "the fraction p over p plus 7 minus 8 equals the fraction 98 over p squared minus 49")[$frac(p, p + 7) − 8 = frac(98, p^(2) − 49)$]

#math.equation(block: true, alt: "the fraction q over 3 q minus 9 minus the fraction 3 over 4 q plus 12 equals the fraction 7 q squared plus 6 q plus 63 over 24 q squared minus 216")[$frac(q, 3 q − 9) − frac(3, 4 q + 12) = frac(7 q^(2) + 6 q + 63, 24 q^(2) − 216)$]

#solutionbox[
#math.equation(block: true, alt: "no solution")[$"no solution"$]
]

#math.equation(block: true, alt: "the fraction r over 3 r minus 15 minus the fraction 1 over 4 r plus 20 equals the fraction 3 r squared plus 17 r plus 40 over 12 r squared minus 300")[$frac(r, 3 r − 15) − frac(1, 4 r + 20) = frac(3 r^(2) + 17 r + 40, 12 r^(2) − 300)$]

#math.equation(block: true, alt: "the fraction s over 2 s plus 6 minus the fraction 2 over 5 s plus 5 equals the fraction 5 s squared minus 3 s minus 7 over 10 s squared plus 40 s plus 30")[$frac(s, 2 s + 6) − frac(2, 5 s + 5) = frac(5 s^(2) − 3 s − 7, 10 s^(2) + 40 s + 30)$]

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

#math.equation(block: true, alt: "the fraction t over 6 t minus 12 minus the fraction 5 over 2 t plus 10 equals the fraction t squared minus 23 t plus 70 over 12 t squared plus 36 t minus 120")[$frac(t, 6 t − 12) − frac(5, 2 t + 10) = frac(t^(2) − 23 t + 70, 12 t^(2) + 36 t − 120)$]

#math.equation(block: true, alt: "the fraction 2 over x squared plus 2 x minus 8 minus the fraction 1 over x squared plus 9 x plus 20 equals the fraction 4 over x squared plus 3 x minus 10")[$frac(2, x^(2) + 2 x − 8) − frac(1, x^(2) + 9 x + 20) = frac(4, x^(2) + 3 x − 10)$]

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

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

#math.equation(block: true, alt: "the fraction 3 over x squared minus 5 x minus 6 plus the fraction 3 over x squared minus 7 x plus 6 equals the fraction 6 over x squared minus 1")[$frac(3, x^(2) − 5 x − 6) + frac(3, x^(2) − 7 x + 6) = frac(6, x^(2) − 1)$]

#solutionbox[
no solution
]

#math.equation(block: true, alt: "the fraction 2 over x squared plus 2 x minus 3 plus the fraction 3 over x squared plus 4 x plus 3 equals the fraction 6 over x squared minus 1")[$frac(2, x^(2) + 2 x − 3) + frac(3, x^(2) + 4 x + 3) = frac(6, x^(2) − 1)$]

#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 minus 2 over x squared plus 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 5")[$f ( x ) = 5$] ⓒ 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 −3 , x equals minus the fraction 14 over 5")[$x = −3 , x = − frac(14, 5)$] ⓒ #math.equation(block: false, alt: "open parenthesis −3 , 5 close parenthesis , open parenthesis minus the fraction 14 over 5 , 5 close parenthesis")[$( −3 , 5 ) , ( − frac(14, 5) , 5 )$]
]

For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x plus 1 over x squared minus 2 x minus 3 ,")[$f ( x ) = frac(x + 1, x^(2) − 2 x − 3) ,$] ⓐ 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.

For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 minus x over x squared minus 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.

#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 5 .")[$x ≠ 5 .$] ⓑ #math.equation(block: false, alt: "x equals the fraction 9 over 2 ,")[$x = frac(9, 2) ,$] ⓒ #math.equation(block: false, alt: "open parenthesis the fraction 9 over 2 , 2 close parenthesis")[$( frac(9, 2) , 2 )$]
]

For rational function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 5 minus x over x squared plus 5 x plus 6 ,")[$f ( x ) = frac(5 − x, x^(2) + 5 x + 6) ,$] #linebreak() ⓐ find the domain of the function #linebreak() ⓑ solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3")[$f ( x ) = 3$] #linebreak() ⓒ the points on the graph at this function value.

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

In the following exercises, solve.

#math.equation(block: false, alt: "the fraction C over r equals 2 π")[$frac(C, r) = 2 π$] for #math.equation(block: false, alt: "r .")[$r .$]

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

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

#math.equation(block: false, alt: "the fraction v plus 3 over w minus 1 equals the fraction 1 over 2")[$frac(v + 3, w − 1) = frac(1, 2)$] for #math.equation(block: false, alt: "w .")[$w .$]

#solutionbox[
#math.equation(block: true, alt: "w equals 2 v plus 7")[$w = 2 v + 7$]
]

#math.equation(block: false, alt: "the fraction x plus 5 over 2 minus y equals the fraction 4 over 3")[$frac(x + 5, 2 − y) = frac(4, 3)$] for #math.equation(block: false, alt: "y .")[$y .$]

#math.equation(block: false, alt: "a equals the fraction b plus 3 over c minus 2")[$a = frac(b + 3, c − 2)$] for #math.equation(block: false, alt: "c .")[$c .$]

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

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

#math.equation(block: false, alt: "the fraction 1 over p plus the fraction 2 over q equals 4")[$frac(1, p) + frac(2, q) = 4$] for #math.equation(block: false, alt: "p .")[$p .$]

#solutionbox[
#math.equation(block: true, alt: "p equals the fraction q over 4 q minus 2")[$p = frac(q, 4 q − 2)$]
]

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

#math.equation(block: false, alt: "the fraction 2 over v plus the fraction 1 over 5 equals the fraction 3 over w")[$frac(2, v) + frac(1, 5) = frac(3, w)$] for #math.equation(block: false, alt: "w .")[$w .$]

#solutionbox[
#math.equation(block: true, alt: "w equals the fraction 15 v over 10 plus v")[$w = frac(15 v, 10 + v)$]
]

#math.equation(block: false, alt: "the fraction 6 over x plus the fraction 2 over 3 equals the fraction 1 over y")[$frac(6, x) + frac(2, 3) = frac(1, y)$] for #math.equation(block: false, alt: "y .")[$y .$]

#math.equation(block: false, alt: "the fraction m plus 3 over n minus 2 equals the fraction 4 over 5")[$frac(m + 3, n − 2) = frac(4, 5)$] for #math.equation(block: false, alt: "n .")[$n .$]

#solutionbox[
#math.equation(block: true, alt: "n equals the fraction 5 m plus 23 over 4")[$n = frac(5 m + 23, 4)$]
]

#math.equation(block: false, alt: "r equals the fraction s over 3 minus t")[$r = frac(s, 3 − t)$] for #math.equation(block: false, alt: "t .")[$t .$]

#math.equation(block: false, alt: "the fraction E over c equals m squared")[$frac(E, c) = m^(2)$] for #math.equation(block: false, alt: "c .")[$c .$]

#solutionbox[
#math.equation(block: true, alt: "c equals the fraction E over m squared")[$c = frac(E, m^(2))$]
]

#math.equation(block: false, alt: "the fraction R over T equals W")[$frac(R, T) = W$] for #math.equation(block: false, alt: "T .")[$T .$]

#math.equation(block: false, alt: "the fraction 3 over x minus the fraction 5 over y equals the fraction 1 over 4")[$frac(3, x) − frac(5, y) = frac(1, 4)$] for #math.equation(block: false, alt: "y .")[$y .$]

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 20 x over 12 minus x")[$y = frac(20 x, 12 − x)$]
]

#math.equation(block: false, alt: "c equals the fraction 2 over a plus the fraction b over 5")[$c = frac(2, a) #h(0.2em) + frac(b, 5)$] for #math.equation(block: false, alt: "a .")[$a .$]


==== Writing Exercises

Your class mate is having trouble in this section. Write down the steps you would use to explain how to solve a rational equation.

#solutionbox[
Answers will vary.
]

Alek thinks the equation #math.equation(block: false, alt: "the fraction y over y plus 6 equals the fraction 72 over y squared minus 36 plus 4")[$frac(y, y + 6) = frac(72, y^(2) − 36) + 4$] has two solutions, #math.equation(block: false, alt: "y equals −6")[$y = −6$] and #math.equation(block: false, alt: "y equals 4 .")[$y = 4 .$] Explain why Alek is wrong.


==== 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 four 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 equations. In row 3, the I can was solve rational equations involving functions. In row 4, the I can was solve rational equations for a specific variable.], alt: "This table has four columns and four 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 equations. In row 3, the I can was solve rational equations involving functions. In row 4, the I can was solve rational equations for a specific variable.", caption: none)

ⓑ On a scale of #math.equation(block: false, alt: "1 minus 10 ,")[$1 − 10 ,$] how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?