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== 7.5#h(0.6em)Equations That Include Algebraic Fractions


    When working with rational functions, we often need to solve equations that involve algebraic fractions.


    In Example of Graphing Rational Functions, we wrote a function that gave the time Francine needs for a 60-mile training run on her cycle-plane in terms of the windspeed, #math.equation(block: false, alt: "x")[$x$]:


    #math.equation(block: true, alt: "t equals f open parenthesis x close parenthesis equals the fraction 60 over 15 minus x")[$t = f ( x ) = frac(60, 15 − x)$]
    If it takes Francine #math.equation(block: false, alt: "9")[$9$] hours to cover #math.equation(block: false, alt: "60")[$60$] miles, what is the speed of the wind? We can answer this question by reading values from the graph of #math.equation(block: false, alt: "f")[$f$], as shown at right. When #math.equation(block: false, alt: "t equals 9")[$t = 9$], the value of #math.equation(block: false, alt: "x")[$x$] is between #math.equation(block: false, alt: "8")[$8$] and #math.equation(block: false, alt: "9")[$9$], so the windspeed is between #math.equation(block: false, alt: "8")[$8$] and #math.equation(block: false, alt: "9")[$9$] miles per hour.


    #figure(figph[graph of rational function for solving equation], alt: "graph of rational function for solving equation", caption: none)


    
      
=== Solving Equations with Fractions Algebraically


      If we need a more accurate value for the windspeed, we can solve the equation


      #math.equation(block: true, alt: "the fraction 60 over 15 minus x equals 9")[$frac(60, 15 − x) = 9$]
      To solve an equation involving an algebraic fraction, we multiply each side of the equation by the denominator of the fraction. This has the effect of clearing the fraction, giving us an equivalent equation without fractions.


      {"equation":"x^2 = 2\*(x + 4)","variable":"x"}-2, 4The method this sentence describes, checked by machine: multiplying x²/(x + 4) = 2 through by the denominator gives x² = 2(x + 4), and the CAS solves that cleared equation at build time, certifying the book's answers x = −2 and x = 4. Neither root makes the original denominator zero, so both survive the check the section teaches next — no extraneous solutions here.#examplebox("Example 1")[][
Solve the equation #math.equation(block: false, alt: "the fraction 60 over 15 minus x equals 9")[$" " " " display(frac(60, 15 − x) = 9)$]


        
        #solutionbox[
We multiply both sides of the equation by #math.equation(block: false, alt: "15 minus x")[$15 − bold(italic(x))$] to obtain


          #math.equation(block: true, alt: "open parenthesis 15 minus x close parenthesis the fraction 60 over 15 minus x, equals 9 open parenthesis 15 minus x close parenthesis; 60, equals 9 open parenthesis 15 minus x close parenthesis, Apply the distributive law.")[$( 15 − bold(italic(x)) ) frac(60, 15 − x) & = 9 ( 15 − bold(italic(x)) ) \ 60 & = 9 ( 15 − x ) &  & "Apply the distributive law."$]
          From here we can proceed as usual.


          #math.equation(block: true, alt: "60, equals 135 minus 9 x, Subtract 135 from both sides.; minus 75, equals minus 9 x, Divide by minus 9.; 8.3 bar, equals x")[$60 & = 135 − 9 x &  & "Subtract 135 from both sides." \ − 75 & = − 9 x &  & "Divide by " − 9. \ 8. limits(3)^(―) & = x$]
          The windspeed was #math.equation(block: false, alt: "8.3 bar")[$8. limits(3)^(―)$], or #math.equation(block: false, alt: "8 the fraction 1 over 3")[$8 frac(1, 3)$] miles per hour.
]
]


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


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different values with a comma.


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


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


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


      
      If the equation contains more than one fraction, we can clear all the denominators at once by multiplying both sides by the LCD of the fractions.


      #examplebox("Example 2")[][
Rani times herself as she kayaks 30 miles down the Derwent River with the help of the current. Returning upstream against the current, she manages only 18 miles in the same amount of time. Rani knows that she can kayak at a rate of 12 miles per hour in still water. What is the speed of the current?


        
        #solutionbox[
If we let #math.equation(block: false, alt: "x")[$x$] represent the speed of the current, we can use the formula #math.equation(block: false, alt: "time equals the fraction distance over rate")[$"time" = display(frac("distance", "rate"))$] to fill in the following table.


          #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [Distance], [Rate], [Time]),
  [Downstream], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "12 plus x")[$12 + x$]], [#math.equation(block: false, alt: "the fraction 30 over 12 plus x")[$display(frac(30, 12 + x))$]],
  [Upstream], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "12 minus x")[$12 − x$]], [#math.equation(block: false, alt: "the fraction 18 over 12 minus x")[$display(frac(18, 12 − x))$]],
))


          Because Rani paddled for equal amounts of time upstream and downstream, we have the equation


          #math.equation(block: true, alt: "the fraction 30 over 12 plus x equals the fraction 18 over 12 minus x")[$frac(30, 12 + x) = frac(18, 12 − x)$]
          The LCD for the fractions in this equation is #math.equation(block: false, alt: "open parenthesis 12 plus x close parenthesis open parenthesis 12 minus x close parenthesis")[$( 12 + x ) ( 12 − x )$]. We multiply both sides of the equation by the LCD to obtain


          #math.equation(block: true, alt: "open parenthesis 12 plus x close parenthesis open parenthesis 12 minus x close parenthesis the fraction 30 over 12 plus x, equals the fraction 18 over 12 minus x open parenthesis 12 plus x close parenthesis open parenthesis 12 minus x close parenthesis; 30 open parenthesis 12 minus x close parenthesis, equals 18 open parenthesis 12 plus x close parenthesis")[$( 12 + bold(italic(x)) ) ( 12 − bold(italic(x)) ) frac(30, 12 + x) & = frac(18, 12 − x) ( 12 + bold(italic(x)) ) ( 12 − bold(italic(x)) ) \ 30 ( 12 − x ) & = 18 ( 12 + x )$]
          Solving this equation, we find


          #math.equation(block: true, alt: "360 minus 30 x, equals 216 plus 18 x; 144, equals 48 x; 3, equals x")[$360 − 30 x & = 216 + 18 x \ 144 & = 48 x \ 3 & = x$]
          The speed of the current is #math.equation(block: false, alt: "3")[$3$] miles per hour.
]
]


      
        Solve #math.equation(block: false, alt: "the fraction x over 6 minus x equals the fraction 1 over 2")[$display(frac(x, 6 − x) = frac(1, 2))$]


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different values with a comma.


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


      
      
        Solve #math.equation(block: false, alt: "the fraction x over 6 minus x equals the fraction 1 over 2")[$display(frac(x, 6 − x) = frac(1, 2))$]


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


      
      
        If the variable in an equation appears in the denominator of a fraction, what should you do to solve the equation?


        \_\_\_\_\_


      
      
        If the variable in an equation appears in the denominator of a fraction, what should you do to solve the equation?


      
      #notebox("Technology", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solving Equations with Fractions Graphically]


        We can solve the equation in Example graphically by considering two functions, one for each side of the equation. Graph the two functions


        #math.equation(block: true, alt: "Y sub 1 equals the fraction 30 over 12 plus x and Y sub 2 equals the fraction 30 over 12 minus x")[$Y_(1) = frac(30, 12 + x) #hide($s p a c e$) "and" #hide($s p a c e$) Y_(2) = frac(30, 12 − x)$]
        in the window


        #math.equation(block: true, alt: "Xmin, equals minus 9.4, Xmax equals 9.4; Ymin, equals 0, Ymax equals 10")[$"Xmin" & = − 9.4 &  & "Xmax" = 9.4 \ "Ymin" & = 0 &  & "Ymax" = 10$]
        to obtain the graph shown below.


        #figure(figph[A graphing calculator screen showing two curves that cross at a single point marked with an X. The calculator's intersect feature reads X = 3, Y = 2. The viewing window runs from negative 9.4 to 9.4 horizontally and up to 10 vertically.], alt: "A graphing calculator screen showing two curves that cross at a single point marked with an X. The calculator's intersect feature reads X = 3, Y = 2. The viewing window runs from negative 9.4 to 9.4 horizontally and up to 10 vertically.", caption: none)


        The function #math.equation(block: false, alt: "Y sub 1")[$Y_(1)$] gives the time it takes Rani to paddle 30 miles downstream, and #math.equation(block: false, alt: "Y sub 2")[$Y_(2)$] gives the time it takes her to paddle #math.equation(block: false, alt: "18")[$18$] miles upstream. Both of these times depend on the speed of the current, #math.equation(block: false, alt: "x")[$x$].


        We are looking for a value of #math.equation(block: false, alt: "x")[$x$] that makes #math.equation(block: false, alt: "Y sub 1")[$Y_(1)$] and #math.equation(block: false, alt: "Y sub 2")[$Y_(2)$] equal. This occurs at the intersection point of the two graphs, #math.equation(block: false, alt: "open parenthesis 3 , 2 close parenthesis")[$( 3 , 2 )$]. Thus, the speed of the current is #math.equation(block: false, alt: "3")[$3$] miles per hour, as we found in Example. The #math.equation(block: false, alt: "y")[$y$]-coordinate of the intersection point gives the time Rani paddled on each part of her trip: #math.equation(block: false, alt: "2")[$2$] hours each way.
]


    
    
      
=== Extraneous Solutions


      A rational function is undefined for any values of #math.equation(block: false, alt: "x")[$x$] that make its denominator equal zero. These values are not in the domain of the function, and they therefore cannot be solutions to equations involving the function. Consider the equation


      #math.equation(block: true, alt: "the fraction x over x minus 3 equals the fraction 3 over x minus 3 plus 2")[$frac(x, x − 3) = frac(3, x − 3) + 2$]
      When we multiply both sides by the LCD, #math.equation(block: false, alt: "x minus 3")[$x − 3$], we obtain


      #math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis the fraction x over x minus 3 equals open parenthesis x minus 3 close parenthesis the fraction 3 over x minus 3 plus open parenthesis x minus 3 close parenthesis times 2")[$( bold(italic(x)) − 3 ) frac(x, x − 3) = ( bold(italic(x)) − 3 ) frac(3, x − 3) + ( bold(italic(x)) − 3 ) ⋅ 2$]
      or


      #math.equation(block: true, alt: "x equals 3 plus 2 x minus 6")[$x = 3 + 2 x − 6$]
      whose solution is


      #math.equation(block: true, alt: "x equals 3")[$x = 3$]
      However, #math.equation(block: false, alt: "x equals 3")[$x = 3$] is not a solution of the original equation. Both sides of the equation are undefined at #math.equation(block: false, alt: "x equals 3")[$x = 3$]. If you graph the two functions


      #math.equation(block: true, alt: "Y sub 1 equals the fraction x over x minus 3 and Y sub 2 equals the fraction 3 over x minus 3 plus 2")[$Y_(1) = frac(x, x − 3) #hide($s p a c e$) "and" #hide($s p a c e$) Y_(2) = frac(3, x − 3) + 2$]
      you will find that the graphs never intersect, which means that there is no solution to the original equation.


      What went wrong with our method of solution? We multiplied both sides of the equation by #math.equation(block: false, alt: "x minus 3")[$x − 3$], which is zero when #math.equation(block: false, alt: "x equals 3")[$x = 3$], so we really multiplied both sides of the equation by zero. Multiplying by zero does not produce an equivalent equation, and false solutions may be introduced.


      An apparent solution that does not satisfy the original equation is called an #strong[extraneous solution]. Whenever we multiply an equation by an expression containing the variable, we should check that the solution obtained is not excluded from the domain of the rational functions involved.


      {"equation":"x = 3 + 2\*x - 6","variable":"x"}3The cautionary example multiplies x/(x − 3) = 3/(x − 3) + 2 through by x − 3 and reaches the cleared equation x = 3 + 2x − 6. The CAS confirms at build time that this cleared equation has exactly one solution, x = 3 — the apparent solution the section then shows FAILING in the original equation, where both sides are undefined at x = 3. The algebra of the clearing step is verified; the lesson is that the clearing step itself, multiplying by zero when x = 3, is what smuggled the false root in.When solving an equation with fractions algebraically, we must be careful to multiply #emph[each] term of the equation by the LCD, no matter whether each term involves fractions.


      #examplebox("Example 3")[][
+ Solve the equation #math.equation(block: false, alt: "the fraction 6 over x plus 1 equals the fraction 1 over x plus 2")[$display(frac(6, x) + 1 = frac(1, x + 2))$] algebraically.
+ Solve the same equation graphically.


        
        #solutionbox[
+ To solve the equation algebraically, we multiply both sides by the LCD, #math.equation(block: false, alt: "x open parenthesis x plus 2 close parenthesis")[$x ( x + 2 )$]. Notice that we multiply each term on the left side by the LCD, to get
				#math.equation(block: true, alt: "x open parenthesis x plus 2 close parenthesis open parenthesis the fraction 6 over x plus 1 close parenthesis equals x open parenthesis x plus 2 close parenthesis the fraction 1 over x plus 2")[$bold(italic(x)) ( bold(italic(x)) + 2 ) ( frac(6, x) + 1 ) = bold(italic(x)) ( bold(italic(x)) + 2 ) frac(1, x + 2)$]
	or
				#math.equation(block: true, alt: "6 open parenthesis x plus 2 close parenthesis plus x open parenthesis x plus 2 close parenthesis equals x")[$6 ( x + 2 ) + x ( x + 2 ) = x$]
	We use the distributive law to remove the parentheses and write the result in standard form:
				#math.equation(block: true, alt: "6 x plus 12 plus x squared plus 2 x, equals x; x squared plus 7 x plus 12, equals 0")[$6 x + 12 + x^(2) + 2 x & = x \ x^(2) + 7 x + 12 & = 0$]
	This is a quadratic equation that we can solve by factoring.
				#math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis open parenthesis x plus 4 close parenthesis equals 0")[$( x + 3 ) ( x + 4 ) = 0$]
	so the solutions are #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$] and #math.equation(block: false, alt: "x equals minus 4")[$x = − 4$]. Neither of these values causes either denominator to equal zero, so they are not extraneous solutions.
+ To solve the equation graphically, graph the two functions
			#math.equation(block: true, alt: "Y sub 1 equals the fraction 6 over x plus 1 and Y sub 2 equals the fraction 1 over x plus 2")[$Y_(1) = frac(6, x) + 1 #hide($s p a c e$) "and" #hide($s p a c e$) Y_(2) = frac(1, x + 2)$]
	in the window
			#math.equation(block: true, alt: "Xmin, equals minus 4.7, Xmax equals 4.7; Ymin, equals minus 10, Ymax equals 10")[$"Xmin" & = − 4.7 &  & "Xmax" = 4.7 \ "Ymin" & = − 10 &  & "Ymax" = 10$]
	as shown in figure (a).#figure(figph[GC images of intersecting rational functions], alt: "GC images of intersecting rational functions", caption: none)

 #linebreak() We see that the first graph has an asymptote at #math.equation(block: false, alt: "x equals 0")[$x = 0$], and the second graph has one at #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$]. It appears that the two graphs may intersect in the third quadrant, around #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$]. To investigate further, we change the window settings to
			#math.equation(block: true, alt: "Xmin, equals minus 4.55, Xmax equals minus 2.2; Ymin, equals minus 1.3, Ymax equals minus 0.3")[$"Xmin" & = − 4.55 &  & "Xmax" = − 2.2 \ "Ymin" & = − 1.3 &  & "Ymax" = − 0.3$]
	to obtain the close-up view shown in figure (b). In this window, we can see that the graphs intersect in two distinct points, and by using the Trace we find that their #math.equation(block: false, alt: "x")[$x$]-coordinates are #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$] and #math.equation(block: false, alt: "x equals minus 4")[$x = − 4$].
]
]


      
        Which of these is the correct first step to solve #math.equation(block: false, alt: "the fraction 1 over x plus the fraction 1 over 2 x plus 1 equals 2")[$display(frac(1, x)) + display(frac(1, 2 x + 1)) = 2$] ?


        \_\_\_\_\_


        #solutionbox[
If we multiply both sides of the equation by the LCD #math.equation(block: false, alt: "x open parenthesis 2 x plus 1 close parenthesis")[$x ( 2 x + 1 )$], we obtain #math.equation(block: false, alt: "2 x plus 1 plus x equals 2 x open parenthesis 2 x plus 1 close parenthesis")[$2 x + 1 + x = 2 x ( 2 x + 1 )$].
]


      
      
        Which of these is the correct first step to solve #math.equation(block: false, alt: "the fraction 1 over x plus the fraction 1 over 2 x plus 1 equals 2")[$display(frac(1, x)) + display(frac(1, 2 x + 1)) = 2$] ?


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


      
      
        Solve #math.equation(block: false, alt: "the fraction 9 over x squared plus x minus 2 plus the fraction 1 over x squared minus 4 equals the fraction 4 over x minus 1")[$display(frac(9, x^(2) + x − 2) + frac(1, x^(2) − 4) = frac(4, x − 1))$]


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Separate different values with a comma.


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


      
      
        Solve #math.equation(block: false, alt: "the fraction 9 over x squared plus x minus 2 plus the fraction 1 over x squared minus 4 equals the fraction 4 over x minus 1")[$" " display(frac(9, x^(2) + x − 2) + frac(1, x^(2) − 4) = frac(4, x − 1))$]


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


      
      
        What are extraneous solutions, and when might they occur?


        \_\_\_\_\_


      
      
        What are extraneous solutions, and when might they occur?


      
    
    
      
=== Formulas


      Algebraic fractions may appear in formulas that relate several variables. If we want to solve for one variable in terms of the others, we may need to clear the fractions.


      #examplebox("Example 4")[][
Solve the formula #math.equation(block: false, alt: "p equals the fraction v over q plus v")[$" " " " p = display(frac(v, q + v) " " " ")$] for #math.equation(block: false, alt: "v")[$v$].


        
        #solutionbox[
Because the variable we want appears in the denominator, we must first multiply both sides of the equation by that denominator, #math.equation(block: false, alt: "q plus v")[$q + v$].


          #math.equation(block: true, alt: "open parenthesis q plus v close parenthesis p, equals open parenthesis q plus v close parenthesis the fraction v over q plus v; open parenthesis q plus v close parenthesis p, equals v")[$( bold(italic(q)) + bold(italic(v)) ) p & = ( bold(italic(q)) + bold(italic(v)) ) frac(v, q + v) \ ( q + v ) p & = v$]
          We apply the distributive law on the left side, then collect all terms that involve #math.equation(block: false, alt: "v")[$v$] on one side of the equation.


          #math.equation(block: true, alt: "q p plus v p, equals v, Subtract v p from both sides.; q p equals v minus v p")[$q p + v p & = v &  & "Subtract " bold(italic(v)) bold(italic(p)) " from both sides." \ q p = v − v p$]
          We cannot combine the two terms containing #math.equation(block: false, alt: "v")[$v$] because they are not like terms. However, we can factor out #math.equation(block: false, alt: "v")[$v$], so that the right side is written as a single term containing the variable #math.equation(block: false, alt: "v")[$v$]. We can then complete the solution.


          #math.equation(block: true, alt: "q p, equals v open parenthesis 1 minus p close parenthesis, Divide both sides by 1 minus p .; the fraction q p over 1 minus p, equals v")[$q p & = v ( 1 − p ) &  & "Divide both sides by " 1 − bold(italic(p)) . \ frac(q p, 1 − p) & = v$]
]
]


      
        Solve for #math.equation(block: false, alt: "a")[$a$]: #math.equation(block: false, alt: "the fraction 2 a b over a plus b equals H")[$#hide($00$) display(frac(2 a b, a + b) = H)$]


        #math.equation(block: false, alt: "a equals")[$a =$]\_\_\_\_\_


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


      
      
        Solve for #math.equation(block: false, alt: "a")[$a$]: #math.equation(block: false, alt: "the fraction 2 a b over a plus b equals H")[$#hide($00$) display(frac(2 a b, a + b) = H)$]


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


      
      
        When you are solving a formula for #math.equation(block: false, alt: "h")[$h$], what should you do if two terms contain #math.equation(block: false, alt: "h")[$h$] as a factor?


        \_\_\_\_\_


      
      
        When you are solving a formula for #math.equation(block: false, alt: "h")[$h$], what should you do if two terms contain #math.equation(block: false, alt: "h")[$h$] as a factor?


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Extraneous solution


      
      
        
==== CONCEPTS


        + To solve an equation involving an algebraic fraction, we multiply each side of the equation by the denominator of the fraction. This has the effect of clearing the fraction, giving us an equivalent equation without fractions.
+ Whenever we multiply an equation by an expression containing the variable, we should check that the solutions obtained are not extraneous.


      
      
        
==== STUDY QUESTIONS


        + What is the first step in solving an equation that includes algebraic fractions?
+ If the equation also contains terms without fractions, should you multiply those terms by the LCD?
+ What are extraneous solutions, and when might they arise?
+ If you are solving a formula and two or more terms contain the variable you are solving for, what should you do?


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Solve a fractional equation by clearing denominators: \#1–14, 47–54
+ Write and solve proportions: \#25–36
+ Solve equations by graphing: \#15–22
+ Solve formulas that involve fractions: \#39–48
+ Solve problems that involve algebraic fractions: \#55–58


      
    
    
      
=== Homework 7.5


      For Problems 1-8, solve the equation algebraically.


      
        #math.equation(block: true, alt: "the fraction 6 over w plus 2 equals 4")[$display(frac(6, w + 2)) = 4$]


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


        
      
      
        #math.equation(block: true, alt: "the fraction 12 over r minus 7 equals 3")[$display(frac(12, r − 7)) = 3$]


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


        
          #math.equation(block: true, alt: "the fraction 13 over 8")[$display(frac(13, 8))$]


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


      
      
        #math.equation(block: true, alt: "the fraction 15 over s squared equals 8")[$display(frac(15, s^(2))) = 8$]


        
          #math.equation(block: true, alt: "plus or minus the square root of the fraction 15 over 8")[$± sqrt(display(frac(15, 8)))$]


        
      
      
        #math.equation(block: true, alt: "the fraction 3 over m squared equals 5")[$display(frac(3, m^(2))) = 5$]


      
      
        #math.equation(block: true, alt: "4.3 equals the square root of the fraction 18 over y")[$4.3 = sqrt(display(frac(18, y)))$]


        
          #math.equation(block: true, alt: "the fraction 1800 over 1849 approximately equals 0.97")[$display(frac(1800, 1849)) ≈ 0.97$]


        
      
      
        #math.equation(block: true, alt: "6.5 equals the fraction 52 over the square root of z")[$6.5 = display(frac(52, sqrt(z)))$]


      
      
        The total weight, #math.equation(block: false, alt: "S")[$S$], that a beam can support is given in pounds by


        #math.equation(block: true, alt: "S equals the fraction 182.6 w h squared over l")[$S = frac(182.6 w h^(2), l)$]
        where #math.equation(block: false, alt: "w")[$w$] is the width of the beam in inches, #math.equation(block: false, alt: "h")[$h$] is its height in inches, and #math.equation(block: false, alt: "l")[$l$] is the length of the beam in feet. A beam over the doorway in an interior wall of a house must support #math.equation(block: false, alt: "1600")[$1600$] pounds. If the beam is #math.equation(block: false, alt: "4")[$4$] inches wide and #math.equation(block: false, alt: "9")[$9$] inches tall, how long can it be?


        
          #math.equation(block: false, alt: "37")[$37$] ft


        
      
      
        If two appliances are connected in parallel in an electrical circuit, the total resistance, #math.equation(block: false, alt: "R")[$R$], in the circuit is given by


        #math.equation(block: true, alt: "R equals the fraction a b over a plus b")[$R = frac(a b, a + b)$]
        where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are the resistances of the two appliances. If one appliance has a resistance of #math.equation(block: false, alt: "18")[$18$] ohms, and the total resistance in the circuit is measured at #math.equation(block: false, alt: "12")[$12$] ohms, what is the resistance of the second appliance?


      
      
        A flock of eider ducks is making a #math.equation(block: false, alt: "150")[$150$]-mile flight at an average airspeed of #math.equation(block: false, alt: "50")[$50$] miles per hour against a moderate
headwind.


        + Express the ducks' travel time, #math.equation(block: false, alt: "t")[$t$], as a function of the windspeed, #math.equation(block: false, alt: "v")[$v$], and graph the function in the window
        #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 50; Ymin, equals 0, Ymax equals 20")[$"Xmin" & = 0 &  & "Xmax" = 50 \ "Ymin" & = 0 &  & "Ymax" = 20$] 
        (See Problem Exercise of Homework Section 7.4.)
+ Write and solve an equation to find the windspeed if the flock makes its trip in #math.equation(block: false, alt: "4")[$4$] hours. Label the corresponding point on your graph.


        
          + #math.equation(block: false, alt: "t equals the fraction 150 over 50 minus v")[$t = display(frac(150, 50 − v))$]#figure(figph[rational function], alt: "rational function", caption: none)
+ #math.equation(block: false, alt: "4 equals the fraction 150 over 50 minus v")[$4 = display(frac(150, 50 − v))$]; #math.equation(block: false, alt: "v equals 12.5")[$v = 12.5$] mph


        
      
      
        Bluefin tuna swim at average speed of #math.equation(block: false, alt: "36")[$36$] miles per hour in still water. A school of tuna is making a #math.equation(block: false, alt: "200")[$200$]-mile trip against a current.


        + Express the tuna's travel time, #math.equation(block: false, alt: "t")[$t$], as a function of the current speed, #math.equation(block: false, alt: "v")[$v$], and graph the function in the window
        #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 36; Ymin, equals 0, Ymax equals 50")[$"Xmin" & = 0 &  & "Xmax" = 36 \ "Ymin" & = 0 &  & "Ymax" = 50$] 
    (See Problem Exercise of Homework Section 7.4.)
+ Write and solve an equation to find the current speed if the school makes its trip in #math.equation(block: false, alt: "8")[$8$] hours. Label the corresponding point on your graph.


      
      
        The cost, in thousands of dollars, for immunizing #math.equation(block: false, alt: "p")[$p$] percent of the residents of Emporia against a dangerous new disease is given by the function


        #math.equation(block: true, alt: "C open parenthesis p close parenthesis equals the fraction 72 p over 100 minus p")[$C ( p ) = frac(72 p, 100 − p)$]
        Write and solve an equation to determine what percent
of the population can be immunized for \$#math.equation(block: false, alt: "168 , 000")[$168 , 000$].


        
          #math.equation(block: true, alt: "168 equals the fraction 72 p over 100 minus p")[$168 = display(frac(72 p, 100 − p))$]; #math.equation(block: true, alt: "p equals 70 %")[$p = 70 upright(%)$]


        
      
      
        The cost, in thousands of dollars, for extracting #math.equation(block: false, alt: "p")[$p$] percent of a precious ore from a mine is given by the function


        #math.equation(block: true, alt: "C open parenthesis p close parenthesis equals the fraction 360 p over 100 minus p")[$C ( p ) = frac(360 p, 100 − p)$]
        Write and solve an equation to determine what percentage of the ore can be extracted for \$#math.equation(block: false, alt: "390 , 000")[$390 , 000$].


      
      For Problems 15–18,


      + Solve the equation graphically by graphing two functions, one for each side of the equation.
+ Solve the equation algebraically.


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


        
          + #figure(figph[rational function and line], alt: "rational function and line", caption: none)
+ #math.equation(block: false, alt: "x equals 1")[$x = 1$]


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


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


        
          + #figure(figph[two rational functions], alt: "two rational functions", caption: none)
+ #math.equation(block: false, alt: "x equals the fraction 1 over 2")[$x = display(frac(1, 2))$]


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


      
      
        The manager of Joe's Burgers discovers that he will sell #math.equation(block: false, alt: "the fraction 160 over x")[$display(frac(160, x))$] burgers per day if the price of a burger is #math.equation(block: false, alt: "x")[$x$] dollars. On the other hand, he can afford to make #math.equation(block: false, alt: "6 x plus 49")[$6 x + 49$] burgers if he charges #math.equation(block: false, alt: "x")[$x$] dollars apiece for them.


        + Graph the #strong[demand function], #math.equation(block: false, alt: "D open parenthesis x close parenthesis equals the fraction 160 over x")[$D ( x ) = display(frac(160, x))$], and the #strong[supply function], #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals 6 x plus 49")[$S ( x ) = 6 x + 49$], in the same window. At what price #math.equation(block: false, alt: "x")[$x$] does the demand for burgers equal the number that Joe can afford to supply? This value for #math.equation(block: false, alt: "x")[$x$] is called the #strong[equilibrium price].
+ Write and solve an equation to verify your equilibrium price.


        
          + #figure(figph[supply and demand curves], alt: "supply and demand curves", caption: none)

 #linebreak() \$#math.equation(block: false, alt: "2.50")[$2.50$]
+ #math.equation(block: false, alt: "the fraction 160 over x equals 6 x plus 49")[$display(frac(160, x)) = 6 x + 49$]; #math.equation(block: false, alt: "x equals 2.50")[$x = 2.50$]


        
      
      
        A florist finds that she will sell #math.equation(block: false, alt: "the fraction 300 over x")[$display(frac(300, x))$] dozen roses per week
if she charges #math.equation(block: false, alt: "x")[$x$] dollars for a dozen. Her suppliers will sell her #math.equation(block: false, alt: "5 x minus 55")[$5 x − 55$] dozen roses if she sells them at #math.equation(block: false, alt: "x")[$x$] dollars per dozen.


        + Graph the demand function, #math.equation(block: false, alt: "D open parenthesis x close parenthesis equals the fraction 300 over x")[$D ( x ) = display(frac(300, x))$], and the supply function, #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals 5 x minus 55")[$S ( x ) = 5 x − 55$], in the same window. At what equilibrium price #math.equation(block: false, alt: "x")[$x$] will the florist sell all the roses she purchases?
+ Write and solve an equation to verify your equilibrium price.


      
      
        Francine wants to fence a rectangular area of #math.equation(block: false, alt: "3200")[$3200$] square feet to grow vegetables for her family of three.


        + Express the length of the garden as a function of its width.
+ Express the perimeter, #math.equation(block: false, alt: "P")[$P$], of the garden as a function of its width.
+ Graph your function for perimeter and find the coordinates of the lowest point on the graph. Interpret those coordinates in the context of the problem.
+ Francine has #math.equation(block: false, alt: "240")[$240$] feet of chain link to make a fence for the garden, and she would like to know what the width of the garden should be. Write an equation that describes this situation.
+ Solve your equation and find the dimensions of the garden.


        
          + #math.equation(block: false, alt: "L equals the fraction 3200 over w")[$L = display(frac(3200, w))$]
+ #math.equation(block: false, alt: "P equals the fraction 6400 over w plus 2 w")[$P = display(frac(6400, w)) + 2 w$]
+ #figure(figph[rational function], alt: "rational function", caption: none)

 #linebreak() Lowest point: #math.equation(block: false, alt: "open parenthesis 56.6 , 226 close parenthesis")[$( 56.6 , 226 )$]; The minimum perimeter is #math.equation(block: false, alt: "226")[$226$] ft for a width of #math.equation(block: false, alt: "56.6")[$56.6$] ft.
+ #math.equation(block: false, alt: "240 equals the fraction 6400 over w plus 2 w")[$240 = display(frac(6400, w)) + 2 w$]
+ #math.equation(block: false, alt: "40")[$40$] ft by #math.equation(block: false, alt: "80")[$80$] ft


        
      
      
        The cost of wire fencing is \$#math.equation(block: false, alt: "7.50")[$7.50$] per foot. A rancher wants to enclose a rectangular pasture of #math.equation(block: false, alt: "1000")[$1000$] square
feet with this fencing.


        + Express the length of the pasture as a function of its width.
+ Express the cost of the fence as a function of its width.
+ Graph your function for the cost and find the coordinates of the lowest point on the graph. Interpret those coordinates in the context of the problem.
+ The rancher has \$#math.equation(block: false, alt: "1050")[$1050$] to spend on the fence, and she would like to know what the width of the pasture should be. Write an equation to describe this situation.
+ Solve your equation and find the dimensions of the pasture.


      
      
        A proportion is an equation in which each side is a ratio: #math.equation(block: false, alt: "the fraction a over b equals the fraction c over d")[$display(frac(a, b)) = display(frac(c, d))$]. Show that this equation may be rewritten as #math.equation(block: false, alt: "a d equals b c")[$a d = b c$].


        
          Multiply both sides of the equation by #math.equation(block: false, alt: "b d")[$b d$] and simplify.


          #math.equation(block: true, alt: "the fraction a over b times the fraction b c over 1 equals the fraction c over d times the fraction b c over 1 , so a c equals b d")[$frac(a, b) ⋅ frac(b c, 1) = frac(c, d) ⋅ frac(b c, 1) , " so " a c = b d$]
        
      
      
        Suppose that #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$], and #math.equation(block: false, alt: "open parenthesis a , b close parenthesis")[$( a , b )$] and
#math.equation(block: false, alt: "open parenthesis c , d close parenthesis")[$( c , d )$] are two points on the graph of #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. Show that #math.equation(block: false, alt: "the fraction b over a equals the fraction d over c")[$display(frac(b, a)) = display(frac(d, c))$].


      
      For Problems 25-28, solve the proportion using your result from Problem 23.


      
        #math.equation(block: true, alt: "the fraction 3 over 4 equals the fraction y plus 2 over 12 minus y")[$display(frac(3, 4)) = display(frac(y + 2, 12 − y))$]


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


        
      
      
        #math.equation(block: true, alt: "the fraction minus 3 over 4 equals the fraction y minus 7 over y plus 14")[$display(frac(− 3, 4)) = display(frac(y − 7, y + 14))$]


      
      
        #math.equation(block: true, alt: "the fraction 50 over r equals the fraction 75 over r plus 20")[$display(frac(50, r)) = display(frac(75, r + 20))$]


        
          #math.equation(block: true, alt: "40")[$40$]


        
      
      
        #math.equation(block: true, alt: "the fraction 30 over r equals the fraction 20 over r minus 10")[$display(frac(30, r)) = display(frac(20, r − 10))$]


      
      For Problems 29-36, use your result from Problem 24 to write and solve a proportion for the problem.


      
        Property taxes on a house vary directly with the value of the house. If the taxes on a house worth \$#math.equation(block: false, alt: "120 , 000")[$120 , 000$] are \$#math.equation(block: false, alt: "2700")[$2700$], what would the taxes be on a house assessed at \$#math.equation(block: false, alt: "275 , 000")[$275 , 000$]?


        
          \$#math.equation(block: false, alt: "6187.50")[$6187.50$]


        
      
      
        The cost of electricity varies directly with the number of units (BTUs) consumed. If a typical household in the Midwest uses #math.equation(block: false, alt: "83")[$83$] million BTUs of electricity annually and pays \$#math.equation(block: false, alt: "1236")[$1236$], how much will a household that uses #math.equation(block: false, alt: "70")[$70$] million BTUs annually spend for energy?


      
      
        Distances on a map vary directly with actual distances. The scale on a map of Michigan uses #math.equation(block: false, alt: "the fraction 3 over 8")[$frac(3, 8)$] inch to represent #math.equation(block: false, alt: "10")[$10$] miles. If Isle Royale is #math.equation(block: false, alt: "1 the fraction 11 over 16")[$1 frac(11, 16)$]
 inches long on the map, what is the actual length of the island?


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


        
      
      
        The dimensions of an enlargement vary directly with the dimensions of the original. A photographer plans to enlarge a photograph that measures #math.equation(block: false, alt: "8.3")[$8.3$] centimeters by #math.equation(block: false, alt: "11.2")[$11.2$] centimeters to produce a poster that is #math.equation(block: false, alt: "36")[$36$] centimeters wide. How long will the poster be?


      
      
        The Forest Service tags #math.equation(block: false, alt: "200")[$200$] perch and releases them into Spirit Lake. One month later, it captures #math.equation(block: false, alt: "80")[$80$] perch and finds that #math.equation(block: false, alt: "18")[$18$] of them are tagged. What is the Forest Service's estimate of the original perch population of the lake?


        
          #math.equation(block: true, alt: "689")[$689$]


        
      
      
        The Wildlife Commission tags #math.equation(block: false, alt: "30")[$30$] Canada geese at one of its migratory feeding grounds. When the geese return, the commission captures #math.equation(block: false, alt: "45")[$45$] geese, of which #math.equation(block: false, alt: "4")[$4$] are tagged. What is the commission's estimate of the number of geese that use the feeding ground?


      
      
        The highest point on Earth is Mount Everest in Tibet, with an elevation of #math.equation(block: false, alt: "8848")[$8848$] meters. The deepest part of the ocean is the Challenger Deep in the Mariana Trench, near Indonesia, #math.equation(block: false, alt: "11 , 034")[$11 , 034$] meters below sea level.


        + What is the total height variation in the surface of the Earth?
+ What percentage of the Earth's radius, #math.equation(block: false, alt: "6400")[$6400$] kilometers, is this variation?
+ If the Earth were shrunk to the size of a basketball, with a radius of #math.equation(block: false, alt: "4.75")[$4.75$] inches, what would be the corresponding height of Mount Everest?


        
          + #math.equation(block: false, alt: "19 , 882")[$19 , 882$] m
+ #math.equation(block: false, alt: "0.3 %")[$0.3 upright(%)$]
+ #math.equation(block: false, alt: "0.00657")[$0.00657$] in


        
      
      
        Shortly after the arrival of human beings at the Hawaiian islands around 400 A.D., many species of birds became extinct. Fossils of #math.equation(block: false, alt: "29")[$29$] different species have been found, but some species may have left no fossils for us to find. We can estimate the total number of extinct species using a proportion. Of #math.equation(block: false, alt: "9")[$9$] species that are still alive, biologists have found fossil evidence of #math.equation(block: false, alt: "7")[$7$]. (Source: Burton, 1998)


        + Assuming that the same fraction of extinct species have left fossil records, calculate the total number of extinct species
+ Give two reasons why this estimate may not be completely accurate.


      
      
        In the figure, the rectangle #math.equation(block: false, alt: "A B C D")[$A B C D$] is divided into a square and a smaller rectangle, #math.equation(block: false, alt: "C D E F")[$C D E F$]. The two rectangles #math.equation(block: false, alt: "A B C D")[$A B C D$] and #math.equation(block: false, alt: "C D E F")[$C D E F$] are similar (their corresponding sides are proportional.) A rectangle #math.equation(block: false, alt: "A B C D")[$A B C D$] with this property is called a #strong[golden rectangle], and the ratio of its length to its width is called the golden ratio.


        The golden ratio appears frequently in art and nature, and it is considered to give the most pleasing proportions to many figures. We will compute the golden ratio as follows.


        #figure(figph[golden rectangle], alt: "golden rectangle", caption: none)


        + Let #math.equation(block: false, alt: "A B equals 1")[$A B = 1$] and #math.equation(block: false, alt: "A D equals x")[$A D = x$]. What are the lengths of #math.equation(block: false, alt: "A E")[$A E$], #math.equation(block: false, alt: "D E")[$D E$], and #math.equation(block: false, alt: "C D")[$C D$]?
+ Write a proportion in terms of #math.equation(block: false, alt: "x")[$x$] for the similarity of rectangles #math.equation(block: false, alt: "A B C D")[$A B C D$] and #math.equation(block: false, alt: "C D E F")[$C D E F$]. Be careful to match up the corresponding sides.
+ Solve your proportion for #math.equation(block: false, alt: "x")[$x$]. Find the golden ratio, #math.equation(block: false, alt: "the fraction A D over A B equals the fraction x over 1")[$display(frac(A D, A B)) = display(frac(x, 1))$].


        
          + #math.equation(block: false, alt: "A E equals 1 , D E equals x minus 1 , C D equals 1")[$A E = 1 , " " D E = x − 1 , " " C D = 1$]
+ #math.equation(block: false, alt: "the fraction 1 over x equals the fraction x minus 1 over x")[$display(frac(1, x)) = display(frac(x − 1, x))$]
+ #math.equation(block: false, alt: "the fraction 1 plus the square root of 5 over 2")[$display(frac(1 + sqrt(5), 2))$]


        
      
      
        The figure shows the graphs of two equations, #math.equation(block: false, alt: "y equals x")[$y = x$] and #math.equation(block: false, alt: "y equals the fraction 1 over x plus 1")[$y = display(frac(1, x)) + 1$].


        + Find the #math.equation(block: false, alt: "x")[$x$]-coordinate of the intersection point of the two graphs.
+ Compare your answer to the golden ratio you computed in Problem 37.


        #figure(figph[line and translated reciprocal], alt: "line and translated reciprocal", caption: none)


      
      For Problems 39-46, solve the formula for the specified variable.


      
        #math.equation(block: false, alt: "S equals the fraction a over 1 minus r")[$S = display(frac(a, 1 − r))$], for #math.equation(block: false, alt: "r")[$r$]


        
          #math.equation(block: true, alt: "r equals the fraction S minus a over S")[$r = display(frac(S − a, S))$]


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


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


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


        
      
      
        #math.equation(block: false, alt: "M equals the fraction a b over a plus b")[$M = display(frac(a b, a + b))$], for #math.equation(block: false, alt: "b")[$b$]


      
      
        #math.equation(block: false, alt: "F equals the fraction G m sub 1 m sub 2 over d squared")[$F = display(frac(G m_(1) m_(2), d^(2)))$], for #math.equation(block: false, alt: "d")[$d$]


        
          #math.equation(block: true, alt: "d equals plus or minus the square root of the fraction G m sub 1 m sub 2 over F")[$d = ± sqrt(display(frac(G m_(1) m_(2), F)))$]


        
      
      
        #math.equation(block: false, alt: "F equals the fraction k q sub 1 q sub 2 over r squared")[$F = display(frac(k q_(1) q_(2), r^(2)))$], for #math.equation(block: false, alt: "r")[$r$]


      
      
        #math.equation(block: false, alt: "the fraction 1 over Q plus the fraction 1 over I equals the fraction 2 over r")[$display(frac(1, Q)) + display(frac(1, I)) = display(frac(2, r))$], for #math.equation(block: false, alt: "r")[$r$]


        
          #math.equation(block: true, alt: "r equals the fraction 2 Q I over I plus Q")[$r = display(frac(2 Q I, I + Q))$]


        
      
      
        #math.equation(block: false, alt: "the fraction 1 over R equals the fraction 1 over A plus the fraction 1 over B")[$display(frac(1, R)) = display(frac(1, A)) + display(frac(1, B))$], for #math.equation(block: false, alt: "B")[$B$]


      
      
        The sidereal period of a planet is the time for the planet to make one trip around the Sun (as seen from the Sun itself). The synodic period is the time between two successive conjunctions of the planet and the Sun, as seen from Earth. The relationship among the sidereal period, #math.equation(block: false, alt: "P")[$P$], of a planet, the synodic period, #math.equation(block: false, alt: "S")[$S$], of the
planet, and the sidereal period of Earth, #math.equation(block: false, alt: "E")[$E$], is given by


        #math.equation(block: true, alt: "the fraction 1 over P equals the fraction 1 over S plus the fraction 1 over E")[$frac(1, P) = frac(1, S) + frac(1, E)$]
        when the planet is closer to the Sun than the Earth is. Solve for #math.equation(block: false, alt: "P")[$P$] in terms of #math.equation(block: false, alt: "S")[$S$] and #math.equation(block: false, alt: "E")[$E$].


        
          #math.equation(block: true, alt: "P equals the fraction E S over E plus S")[$P = display(frac(E S, E + S))$]


        
      
      
        When a planet is farther from the Sun than Earth is,


        #math.equation(block: true, alt: "the fraction 1 over P equals the fraction 1 over E minus the fraction 1 over S")[$frac(1, P) = frac(1, E) − frac(1, S)$]
        where #math.equation(block: false, alt: "P")[$P$], #math.equation(block: false, alt: "E")[$E$], and #math.equation(block: false, alt: "S")[$S$] are as defined in Problem 47. Solve for #math.equation(block: false, alt: "P")[$P$] in terms of #math.equation(block: false, alt: "S")[$S$] and #math.equation(block: false, alt: "E")[$E$].


      
      For Problems 49-56, solve the equation algebraically.


      
        #math.equation(block: true, alt: "the fraction 3 over x minus 2 equals the fraction 1 over 2 plus the fraction 2 x minus 7 over 2 x minus 4")[$display(frac(3, x − 2)) = display(frac(1, 2)) + display(frac(2 x − 7, 2 x − 4))$]


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


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


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


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


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


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


        
          #math.equation(block: true, alt: "the fraction minus 14 over 5")[$display(frac(− 14, 5))$]


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


      
      
        #math.equation(block: true, alt: "the fraction 4 over 3 x plus the fraction 3 over 3 x plus 1 plus 2 equals 0")[$display(frac(4, 3 x)) + display(frac(3, 3 x + 1)) + 2 = 0$]


        
          #math.equation(block: true, alt: "the fraction minus 1 over 6 , the fraction minus 4 over 3")[$display(frac(− 1, 6)) , display(frac(− 4, 3))$]


        
      
      
        #math.equation(block: true, alt: "minus 3 equals the fraction minus 10 over x plus 2 plus the fraction 10 over x plus 5")[$− 3 = display(frac(− 10, x + 2)) + display(frac(10, x + 5))$]


      
      
        A chartered sightseeing flight over the Grand Canyon is scheduled to return to its departure point in #math.equation(block: false, alt: "3")[$3$] hours. The pilot would like to cover a distance of #math.equation(block: false, alt: "144")[$144$] miles before turning around, and he hears on the Weather Service that there will be a headwind of #math.equation(block: false, alt: "20")[$20$] miles per hour on the outward journey.


        + Express the time it takes for the outward journey as a function of the airspeed of the plane.
+ Express the time it takes for the return journey as a function of the speed of the plane.
+ Graph the sum of the two functions and find the point on the graph with #math.equation(block: false, alt: "y")[$y$]-coordinate #math.equation(block: false, alt: "3")[$3$]. Interpret the coordinates of the point in the context of the problem.
+ The pilot would like to know what airspeed to maintain in order to complete the tour in #math.equation(block: false, alt: "3")[$3$] hours. Write an equation to describe this situation.
+ Solve your equation to find the appropriate airspeed.


        
          + #math.equation(block: false, alt: "t sub 1 equals the fraction 144 over s minus 20")[$t_(1) = display(frac(144, s − 20))$]
+ #math.equation(block: false, alt: "t sub 2 equals the fraction 144 over s plus 20")[$t_(2) = display(frac(144, s + 20))$]
+ #figure(figph[rational function], alt: "rational function", caption: none)

 #linebreak() If the airspeed is #math.equation(block: false, alt: "100")[$100$] mph, the round trip will take #math.equation(block: false, alt: "3")[$3$] hours.
+ #math.equation(block: false, alt: "the fraction 144 over s minus 20 plus the fraction 144 over s plus 20 equals 3")[$display(frac(144, s − 20)) + display(frac(144, s + 20)) = 3$]
+ #math.equation(block: false, alt: "100")[$100$] mph


        
      
      
        Two student pilots leave the airport at the same time. They both fly at an airspeed of #math.equation(block: false, alt: "180")[$180$] miles per hour, but one flies with the wind and the other flies against the
wind.


        + Express the time it takes the first pilot to travel #math.equation(block: false, alt: "500")[$500$] miles as a function of the windspeed.
+ Express the time it takes the second pilot to travel #math.equation(block: false, alt: "400")[$400$] miles as a function of the windspeed.
+ Graph the two functions in the same window, and find the coordinates of the intersection point. Interpret those coordinates in the context of the problem.
+ Both pilots check in with their instructors at the same time, and the first pilot has traveled #math.equation(block: false, alt: "500")[$500$] miles while the second pilot has gone #math.equation(block: false, alt: "400")[$400$] miles. Write an equation to describe this situation.
+ Solve your equation to find the speed of the wind.


      
      
        Andy drives #math.equation(block: false, alt: "300")[$300$] miles to Lake Tahoe at #math.equation(block: false, alt: "70")[$70$] miles per hour and returns home at #math.equation(block: false, alt: "50")[$50$] miles per hour. What is his average speed for the round trip? (It is not #math.equation(block: false, alt: "60")[$60$] miles per hour!)


        + Write expressions for the time it takes for each leg of the trip if Andy drives a distance, #math.equation(block: false, alt: "d")[$d$], at speed #math.equation(block: false, alt: "r sub 1")[$r_(1)$] and returns at speed #math.equation(block: false, alt: "r sub 2")[$r_(2)$].
+ Write expressions for the total distance and total time for the trip.
+ Write an expression for the average speed for the entire trip.
+ Write your answer to part (c) as a simple fraction.
+ Use your formula to answer the question stated in the problem.


        
          + #math.equation(block: false, alt: "t sub 1 equals the fraction d over r sub 1 , t sub 2 equals the fraction d over r sub 2")[$t_(1) = display(frac(d, r_(1))) , " " t_(2) = display(frac(d, r_(2)))$]
+ Total distance is #math.equation(block: false, alt: "2 d")[$2 d$]; total time #math.equation(block: false, alt: "the fraction d over r sub 1 plus the fraction d over r sub 2")[$display(frac(d, r_(1))) + display(frac(d, r_(2)))$].
+ #math.equation(block: false, alt: "the fraction 2 d over the fraction d over r sub 1 plus the fraction d over r sub 2")[$display(frac(2 d, display(frac(d, r_(1))) + display(frac(d, r_(2)))))$]
+ #math.equation(block: false, alt: "the fraction 2 r sub 1 r sub 2 over r sub 1 plus r sub 2")[$display(frac(2 r_(1) r_(2), r_(1) + r_(2)))$]
+ #math.equation(block: false, alt: "58 the fraction 1 over 3")[$58 frac(1, 3)$] mph


        
      
      
        The owner of a print shop volunteers to produce flyers for his candidate's campaign. His large printing press can complete the job in #math.equation(block: false, alt: "4")[$4$] hours, and the smaller model can finish the flyers in #math.equation(block: false, alt: "6")[$6$] hours. How long will it take to print the flyers if he runs both presses simultaneously?


        + Suppose that the large press can complete a job in #math.equation(block: false, alt: "t sub 1")[$t_(1)$] hours and the smaller press takes #math.equation(block: false, alt: "t sub 2")[$t_(2)$] hours. Write expressions for the fraction of a job that each press can complete in #math.equation(block: false, alt: "1")[$1$] hour.
+ Write an expression for the fraction of a job that can be completed in #math.equation(block: false, alt: "1")[$1$] hour with both presses running simultaneously.
+ Write an expression for the amount of time needed to complete the job with both presses running.
+ Write your answer to part (c) as a simple fraction.
+ Use your formula to answer the question stated in the problem.