Elementary Algebra 2e · free preview

8.6 Solve Rational Equations

Learning objectives: Solve rational equations; Solve a rational equation for a specific variable

1. Learning objectives

Learning objectives

Learning objectives

  • Solve rational equations
  • Solve a rational equation for a specific variable

After defining the terms expression and equation early in Foundations, we have used them throughout this course. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve rational equations.

The definition of a rational equation is similar to the definition of equation we used in Foundations.

A rational equation is two rational expressions connected by an equal sign.

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

RationalExpression RationalEquation 18x+12 18x+12=14 y+6y2−36 y+6y2−36=y+1 1n−3+1n+4 1n−3+1n+4=15n2+n−12\begin{array}{cccc} \mathrm{Rational Expression} & & & \,\mathrm{Rational Equation} \\ \,\frac{1}{8}x+\frac{1}{2} & & & \,\frac{1}{8}x+\frac{1}{2}=\frac{1}{4} \\ \,\frac{y+6}{{y}^{2}-36} & & & \,\frac{y+6}{{y}^{2}-36}=y+1 \\ \,\frac{1}{n-3}+\frac{1}{n+4} & & & \,\frac{1}{n-3}+\frac{1}{n+4}=\frac{15}{{n}^{2}+n-12} \\ \end{array}

2. Solve Rational Equations

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.

Here is an example we did when we worked with linear equations:

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 extraneous solution.

An 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, c, by writing x≠cx\ne{}c next to the equation.

3. Example 8.59

Example 8.59

Example 8.59

How to Solve Equations with Rational Expressions

Solve: 1x+13=56.\frac{1}{x}+\frac{1}{3}=\frac{5}{6}.

The steps of this method are shown below.

Solve equations with rational expressions.

  1. Step 1. Note any value of the variable that would make any denominator zero.
  2. Step 2. Find the least common denominator of all denominators in the equation.
  3. Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
  4. Step 4. Solve the resulting equation.
  5. Step 5. 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.

4. Example 8.60

Example 8.60

Example 8.60

Solve: 1−5y=−6y2.1-\frac{5}{y}=-\frac{6}{{y}^{2}}.

Solution

5. Example 8.61

Example 8.61

Example 8.61

Solve: 53u−2=32u.\frac{5}{3u-2}=\frac{3}{2u}.

Solution

When one of the denominators is a quadratic, remember to factor it first to find the LCD.

6. Example 8.62

Example 8.62

Example 8.62

Solve: 2p+2+4p−2=p−1p2−4.\frac{2}{p+2}+\frac{4}{p-2}=\frac{p-1}{{p}^{2}-4}.

Solution

7. Example 8.63

Example 8.63

Example 8.63

Solve: 4q−4−3q−3=1.\frac{4}{q-4}-\frac{3}{q-3}=1.

Solution

8. Example 8.64

Example 8.64

Example 8.64

Solve: m+11m2−5m+4=5m−4−3m−1.\frac{m+11}{{m}^{2}-5m+4}=\frac{5}{m-4}-\frac{3}{m-1}.

Solution

The equation we solved in Example 8.64 had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.

9. Example 8.65

Example 8.65

Example 8.65

Solve: n12+n+33n=1n.\frac{n}{12}+\frac{n+3}{3n}=\frac{1}{n}.

Solution

10. Example 8.66

Example 8.66

Example 8.66

Solve: yy+6=72y2−36+4.\frac{y}{y+6}=\frac{72}{{y}^{2}-36}+4.

Solution

11. Example 8.67

Example 8.67

Example 8.67

Solve: x2x−2−23x+3=5x2−2x+912x2−12.\frac{x}{2x-2}-\frac{2}{3x+3}=\frac{5{x}^{2}-2x+9}{12{x}^{2}-12}.

Solution

12. Solve a Rational Equation for a Specific Variable

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.

We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.

13. Example 8.68

Example 8.68

Example 8.68

Solve: DT=R for T.\frac{D}{T}=R\,\mathrm{for}\,T.

Solution

Example 8.69 uses the formula for slope that we used to get the point-slope form of an equation of a line.

14. Example 8.69

Example 8.69

Example 8.69

Solve: m=x−2y−3 for y.m=\frac{x-2}{y-3}\,\mathrm{for}\,y.

Solution

Be sure to follow all the steps in Example 8.70. It may look like a very simple formula, but we cannot solve it instantly for either denominator.

15. Example 8.70

Example 8.70

Example 8.70

Solve 1c+1m=1 for c.\frac{1}{c}+\frac{1}{m}=1\,\mathrm{for}\,c.

Solution

Notice that even though we excluded c=0 and m=0c=0\,\mathrm{and}\,m=0 from the original equation, we must also now state that m≠1m\ne{}1.

16. Section 8.6 Exercises

Section 8.6 Exercises

Section 8.6 Exercises

Solve Rational Equations

In the following exercises, solve.

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

Everyday Math

Writing Exercises

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

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

17. Practice 1

  1. Solve: 1y+23=15.\frac{1}{y}+\frac{2}{3}=\frac{1}{5}.
  1. Solve: 23+15=1x.\frac{2}{3}+\frac{1}{5}=\frac{1}{x}.
  1. Solve: 1−2a=15a2.1-\frac{2}{a}=\frac{15}{{a}^{2}}.
  1. Solve: 1−4b=12b2.1-\frac{4}{b}=\frac{12}{{b}^{2}}.

18. Practice 2

  1. Solve: 1x−1=23x.\frac{1}{x-1}=\frac{2}{3x}.
  1. Solve: 35n+1=23n.\frac{3}{5n+1}=\frac{2}{3n}.
  1. Solve: 2x+1+1x−1=1x2−1.\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{{x}^{2}-1}.
  1. Solve: 5y+3+2y−3=5y2−9.\frac{5}{y+3}+\frac{2}{y-3}=\frac{5}{{y}^{2}-9}.

19. Practice 3

  1. Solve: 2x+5−1x−1=1.\frac{2}{x+5}-\frac{1}{x-1}=1.
  1. Solve: 3x+8−2x−2=1.\frac{3}{x+8}-\frac{2}{x-2}=1.
  1. Solve: x+13x2−7x+10=6x−5−4x−2.\frac{x+13}{{x}^{2}-7x+10}=\frac{6}{x-5}-\frac{4}{x-2}.
  1. Solve: y−14y2+3y−4=2y+4+7y−1.\frac{y-14}{{y}^{2}+3y-4}=\frac{2}{y+4}+\frac{7}{y-1}.

20. Practice 4

  1. Solve: x18+x+69x=23x.\frac{x}{18}+\frac{x+6}{9x}=\frac{2}{3x}.
  1. Solve: y+55y+y15=1y.\frac{y+5}{5y}+\frac{y}{15}=\frac{1}{y}.
  1. Solve: xx+4=32x2−16+5.\frac{x}{x+4}=\frac{32}{{x}^{2}-16}+5.
  1. Solve: yy+8=128y2−64+9.\frac{y}{y+8}=\frac{128}{{y}^{2}-64}+9.

21. Practice 5

  1. Solve: y5y−10−53y+6=2y2−19y+5415y2−60.\frac{y}{5y-10}-\frac{5}{3y+6}=\frac{2{y}^{2}-19y+54}{15{y}^{2}-60}.
  1. Solve: z2z+8−34z−8=3z2−16z−168z2+16z−64.\frac{z}{2z+8}-\frac{3}{4z-8}=\frac{3{z}^{2}-16z-16}{8{z}^{2}+16z-64}.
  1. Solve: AL=W\frac{A}{L}=W for L.L.
  1. Solve: FA=M\frac{F}{A}=M for A.A.

22. Practice 6

  1. Solve: y−2x+1=23\frac{y-2}{x+1}=\frac{2}{3} for x.x.
  1. Solve: x=y1−yx=\frac{y}{1-y} for y.y.
  1. Solve: 1a+1b=c\frac{1}{a}+\frac{1}{b}=c for a.a.
  1. Solve: 2x+13=1y\frac{2}{x}+\frac{1}{3}=\frac{1}{y} for y.y.

23. Practice 7

  1. 1a+25=12\frac{1}{a}+\frac{2}{5}=\frac{1}{2}
  1. 52−1c=34\frac{5}{2}-\frac{1}{c}=\frac{3}{4}
  1. 45+14=2v\frac{4}{5}+\frac{1}{4}=\frac{2}{v}
  1. 79+1x=23\frac{7}{9}+\frac{1}{x}=\frac{2}{3}

24. Practice 8

  1. 1−2m=8m21-\frac{2}{m}=\frac{8}{{m}^{2}}
  1. 1+9p=−20p21+\frac{9}{p}=\frac{-20}{{p}^{2}}
  1. 1r+3=42r\frac{1}{r+3}=\frac{4}{2r}
  1. 53v−2=74v\frac{5}{3v-2}=\frac{7}{4v}

25. Practice 9

  1. 3x+4+7x−4=8x2−16\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{{x}^{2}-16}
  1. 8z−10+7z+10=5z2−100\frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{{z}^{2}-100}
  1. 1q+4−2q−2=1\frac{1}{q+4}-\frac{2}{q-2}=1
  1. 1t+7−5t−5=1\frac{1}{t+7}-\frac{5}{t-5}=1

26. Practice 10

  1. v−10v2−5v+4=3v−1−6v−4\frac{v-10}{{v}^{2}-5v+4}=\frac{3}{v-1}-\frac{6}{v-4}
  1. x−10x2+8x+12=3x+2+4x+6\frac{x-10}{{x}^{2}+8x+12}=\frac{3}{x+2}+\frac{4}{x+6}
  1. z16+z+24z=12z\frac{z}{16}+\frac{z+2}{4z}=\frac{1}{2z}
  1. b+33b+b24=1b\frac{b+3}{3b}+\frac{b}{24}=\frac{1}{b}

27. Practice 11

  1. dd+3=18d2−9+4\frac{d}{d+3}=\frac{18}{{d}^{2}-9}+4
  1. nn+2=8n2−4+3\frac{n}{n+2}=\frac{8}{{n}^{2}-4}+3
  1. q3q−9−34q+12\frac{q}{3q-9}-\frac{3}{4q+12} =7q2+6q+6324q2−216\,=\frac{7{q}^{2}+6q+63}{24{q}^{2}-216}
  1. s2s+6−25s+5\frac{s}{2s+6}-\frac{2}{5s+5} =5s2−s−1810s2+40s+30\,=\frac{5{s}^{2}-s-18}{10{s}^{2}+40s+30}

28. Practice 12

  1. Cr=2π for r\frac{C}{r}=2\pi\,\mathrm{for}\,r
  1. Vh=lw for h\frac{V}{h}=lw\,\mathrm{for}\,h
  1. v+3w−1=12 for w\frac{v+3}{w-1}=\frac{1}{2}\,\mathrm{for}\,w
  1. a=b+3c−2 for ca=\frac{b+3}{c-2}\,\mathrm{for}\,c

29. Practice 13

  1. 1p+2q=4 for p\frac{1}{p}+\frac{2}{q}=4\,\mathrm{for}\,p
  1. 2v+15=3w for w\frac{2}{v}+\frac{1}{5}=\frac{3}{w}\,\mathrm{for}\,w
  1. m+3n−2=45 for n\frac{m+3}{n-2}=\frac{4}{5}\,\mathrm{for}\,n
  1. 3x−5y=14 for y\frac{3}{x}-\frac{5}{y}=\frac{1}{4}\,\mathrm{for}\,y

30. Practice 14

  1. r=s3−t for tr=\frac{s}{3-t}\,\mathrm{for}\,t
  1. House Painting Alain can paint a house in 4 days. Spiro would take 7 days to paint the same house. Solve the equation 14+17=1t\frac{1}{4}+\frac{1}{7}=\frac{1}{t} for t to find the number of days it would take them to paint the house if they worked together.
  1. Why is there no solution to the equation 3x−2=5x−2\frac{3}{x-2}=\frac{5}{x-2}?

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