4.7: Solving Rational Equations

Learning Objectives

• Solve rational equations.
• Solve literal equations, or formulas, involving rational expressions.
• Solve applications involving the reciprocal of unknowns.

Solving Rational Equations

A rational equation³³ is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not \(0\) by making note of restrictions and checking our solutions. Solving rational equations involves clearing fractions by multiplying both sides of the equation by the least common denominator (LCD).

Example \(\PageIndex{1}\):

Solve: \(\frac { 1 } { x } + \frac { 2 } { x ^ { 2 } } = \frac { x + 9 } { 2 x ^ { 2 } }\).

Solution

We first make a note of the restriction on \(x, x≠0\). We then multiply both sides by the LCD, which in this case equals \(2x^{2}\).

\(\begin{aligned} \color{Cerulean}{2 x ^ { 2} }\color{black}{ \cdot} \left( \frac { 1 } { x } + \frac { 2 } { x ^ { 2 } } \right) & =\color{Cerulean}{ 2 x ^ { 2} }\color{black}{ \cdot} \left( \frac { x + 9 } { 2 x ^ { 2 } } \right)\quad\quad\color{Cerulean}{Multiply\:both\:sides\:by\:the\:LCD.} \\ \color{Cerulean}{2 x ^ { 2} }\color{black}{ \cdot} \frac { 1 } { x } + \color{Cerulean}{2 x ^ { 2} }\color{black}{ \cdot} \frac { 2 } { x ^ { 2 } } & = \color{Cerulean}{2 x ^ { 2 }}\color{black}{ \cdot} \frac { x + 9 } { 2 x ^ { 2 } }\quad\quad\quad\:\:\color{Cerulean}{Distribute.} \\ 2 x + 4 & = x + 9\quad\quad\quad\quad\quad\quad\color{Cerulean}{Simplify\:and\:then\:solve.} \\ x & = 5 \end{aligned}\)

Check your answer. Substitute \(x=5\) into the original equation and see if you obtain a true statement.

\(\begin{array} { l } { \frac { 1 } { x } + \frac { 2 } { x ^ { 2 } } = \frac { x + 9 } { 2 x ^ { 2 } }\quad \color{Cerulean} { Original\:equation } } \\ { \frac { 1 } { \color{OliveGreen}{5} }\color{black}{ +} \frac { 2 } { \color{OliveGreen}{5} ^ { \color{black}{2} } } = \frac { \color{OliveGreen}{5}\color{black}{ +} 9 } { 2 ( \color{OliveGreen}{5}\color{black}{ )} ^ { 2 } }\quad \color{Cerulean} { Check \: x = 5.} } \\ { \frac { 1 } { 5 } + \frac { 2 } { 25 } = \frac { 14 } { 2 \cdot 25 } } \\ \frac{5}{25} + \frac{2}{25} = \frac{7}{25} \\ {\frac{7}{25}=\frac{7}{25}} \:\:\color{Cerulean}{✓}\end{array}\)

Answer:

The solution is \(5\).

After multiplying both sides of the previous example by the LCD, we were left with a linear equation to solve. This is not always the case; sometimes we will be left with quadratic equation.

Example \(\PageIndex{2}\):

Solve: \(\frac { 3 ( x + 2 ) } { x - 4 } - \frac { x + 4 } { x - 2 } = \frac { x - 2 } { x - 4 }\).

Solution

In this example, there are two restrictions, \(x≠4\) and \(x≠2\). Begin by multiplying both sides by the LCD, \((x−2)(x−4)\).

\(\begin{aligned} \color{Cerulean}{(x-2)(x-4)}\color{black}{\cdot} \left( \frac{3(x+2)}{x-4}-\frac{x+4}{x-2} \right) &= \color{Cerulean}{(x-2)(x-4)}\color{black}{\cdot} \left( \frac{x-2}{x-4} \right)\\ \color{Cerulean}{(x-2)\cancel{(x-4)}}\color{black}{\cdot} \frac{3(x+2)}{\cancel{x-4}} - \color{Cerulean}{\cancel{(x-2)}(x-4)}\color{black}{\cdot} \frac{x+4}{\cancel{x-2}} &= \color{Cerulean}{(x-2)\cancel{(x-4)}}\color{black}{\cdot} \frac{x-2}{\cancel{x-4}} \\ 3 ( x + 2 ) ( x - 2 ) - ( x + 4 ) ( x - 4 ) & = ( x - 2 ) ( x - 2 ) \\ 3 \left( x ^ { 2 } - 4 \right) - \left( x ^ { 2 } - 16 \right) & = x ^ { 2 } - 2 x - 2 x + 4 \\ 3 x ^ { 2 } - 12 - x ^ { 2 } + 16 & = x ^ { 2 } - 4 x + 4\\ 2x^{2} + 4 & = x^{2}-4x+4 \end{aligned}\)

After distributing and simplifying both sides of the equation, a quadratic equation remains. To solve, rewrite the quadratic equation in standard form, factor, and then set each factor equal to 0.

\(\begin{array} { l } { 2 x ^ { 2 } + 4 = x ^ { 2 } - 4 x + 4 } \\ { x ^ { 2 } + 4 x = 0 } \\ { x ( x + 4 ) = 0 } \end{array}\)

\(\begin{aligned} x = 0 \text { or } x + 4 & = 0 \\ x & = - 4 \end{aligned}\)

Check to see if these values solve the original equation.

\(\frac { 3 ( x + 2 ) } { x - 4 } - \frac { x + 4 } { x - 2 } = \frac { x - 2 } { x - 4 }\)

Table \(\PageIndex{1}\)

Check \(x=0\)

Check \(x=4\)

\(\begin{aligned} \frac { 3 ( \color{Cerulean}{0}\color{black}{ +} 2 ) } {\color{Cerulean}{ 0}\color{black}{ -} 4 } - \frac {\color{Cerulean}{ 0}\color{black}{ +} 4 } {\color{Cerulean}{ 0}\color{black}{ -} 2 } & = \frac { \color{Cerulean}{0}\color{black}{ -} 2 } { \color{Cerulean}{0}\color{black}{ -} 4 } \\ \frac { 6 } { - 4 } - \frac { 4 } { - 2 } & = \frac { - 2 } { - 4 } \\ - \frac { 3 } { 2 } + 2 & = \frac { 1 } { 2 } \\ - \frac { 3 } { 2 } + \frac { 4 } { 2 } & = \frac { 1 } { 2 } \\ \frac { 1 } { 2 } & = \frac { 1 } { 2 } \color{Cerulean}{✓} \end{aligned}\)

\(\begin{aligned} \frac { 3 ( \color{Cerulean}{- 4}\color{black}{ +} 2 ) } { \color{Cerulean}{- 4}\color{black}{ -} 4 } - \frac {\color{Cerulean}{ - 4}\color{black}{ +} 4 } {\color{Cerulean}{ - 4}\color{black}{ -} 2 } & = \frac { \color{Cerulean}{- 4}\color{black}{ -} 2 } { \color{Cerulean}{- 4}\color{black}{ -} 4 } \\ \frac { 3 ( - 2 ) } { - 8 } - \frac { 0 } { - 6 } & = \frac { - 6 } { - 8 } \\ \frac { - 6 } { - 8 } - 0 & = \frac { 3 } { 4 } \\ \frac { 3 } { 4 } & = \frac { 3 } { 4 } \color{Cerulean}{✓} \end{aligned}\)

Answer:

The solutions are \(0\) and \(−4\).

Up to this point, all of the possible solutions have solved the original equation. However, this may not always be the case. Multiplying both sides of an equation by variable factors may lead to extraneous solutions³⁴, which are solutions that do not solve the original equation. A complete list of steps for solving a rational equation is outlined in the following example.

Example \(\PageIndex{3}\)

Solve: \(\frac { 2 x } { 3 x + 1 } = \frac { 1 } { x - 5 } - \frac { 4 ( x - 1 ) } { 3 x ^ { 2 } - 14 x - 5 }\).

Solution

Step 1: Factor all denominators and determine the LCD.

\(\begin{array} { l } { \frac { 2 x } { 3 x + 1 } = \frac { 1 } { x - 5 } - \frac { 4 ( x - 1 ) } { 3 x ^ { 2 } - 14 x - 5 } } \\ { \frac { 2 x } { ( 3 x + 1 ) } = \frac { 1 } { ( x - 5 ) } - \frac { 4 ( x - 1 ) } { ( 3 x + 1 ) ( x - 5 ) } } \end{array}\)

The LCD is \((3x+1)(x−5)\).

Step 2: Identify the restrictions. In this case, \(x≠−\frac{1}{3}\) and \(x≠5\).

Step 3: Multiply both sides of the equation by the LCD. Distribute carefully and then simplify.

\(\begin{aligned} \color{Cerulean}{( 3 x + 1 ) ( x - 5 )}\color{black}{ \cdot} \frac { 2 x } { ( 3 x + 1 ) } &=\color{Cerulean}{ ( 3 x + 1 ) ( x - 5 )}\color{black}{ \cdot} \left( \frac { 1 } { ( x - 5 ) } - \frac { 4 ( x - 1 ) } { ( 3 x + 1 ) ( x - 5 ) }\right)\\ \color{Cerulean}{\cancel{(3x+1)}(x-5)}\color{black}{ \cdot}\frac{2x}{\cancel{(3x+1)}} &= \color{Cerulean}{(3x+1)\cancel{(x-5)}}\color{black}{\cdot} \frac{1}{\cancel{(x-5)}} -\color{Cerulean}{\cancel{ (3x+1)}\cancel{(x-5)}}\color{black}{ \cdot} \frac{4(x-1)}{\cancel{(3x+1)}\cancel{(x-5)}} \\ 2x(x-5) &=(3x+1)-4(x-1) \end{aligned}\)

Step 4: Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to \(0\).

\(\begin{aligned} 2 x ( x - 5 ) & = ( 3 x + 1 ) - 4 ( x - 1 ) \\ 2 x ^ { 2 } - 10 x & = 3 x + 1 - 4 x + 4 \\ 2 x ^ { 2 } - 10 x & = - x + 5 \\ 2 x ^ { 2 } - 9 x - 5 & = 0 \\ ( 2 x + 1 ) ( x - 5 ) & = 0 \end{aligned}\)

\(\begin{aligned} 2 x + 1 & = 0 \quad\quad \text { or } &x - 5 &= 0 \\ 2 x & = - 1 & x &= 5 \\ x &= - \frac { 1 } { 2 } \end{aligned}\)

Step 5: Check for extraneous solutions. Always substitute into the original equation, or the factored equivalent. In this case, choose the factored equivalent to check:

\(\frac { 2 x } { ( 3 x + 1 ) } = \frac { 1 } { ( x - 5 ) } - \frac { 4 ( x - 1 ) } { ( 3 x + 1 ) ( x - 5 ) }\)

Table \(\PageIndex{2}\)

Check \(x=-\frac{1}{2}\)	Check \(x=5\)
\(\begin{aligned} \frac { 2 \left( \color{Cerulean}{- \frac { 1 } { 2 }} \right) } { \left( 3 \left(\color{Cerulean}{ - \frac { 1 } { 2 }} \right) \color{black}{+} 1 \right) } & = \frac { 1 } { \left( \left( \color{Cerulean}{- \frac { 1 } { 2} } \right) - 5 \right) } - \frac { 4 \left( \left( \color{Cerulean}{- \frac { 1 } { 2 }} \right) \color{black}{-} 1 \right) } { \left( 3 \left( \color{Cerulean}{- \frac { 1 } { 2} } \right) \color{black}{+} 1 \right) \left( \left( \color{Cerulean}{- \frac { 1 } { 2} } \right) - 5 \right) } \\ \frac { - 1 } { \left( - \frac { 1 } { 2 } \right) } & = \frac { 1 } { \left( - \frac { 11 } { 2 } \right) } - \frac { 4 \left( - \frac { 3 } { 2 } \right) } { \left( - \frac { 1 } { 2 } \right) \left( - \frac { 11 } { 2 } \right) } \\ 2 & = -\frac{2}{11} - \frac{-6}{\left(\frac{11}{4} \right)} \\ 2 & = - \frac { 2 } { 11 } + \frac { 24 } { 11 } \\ 2 & = \frac { 22 } { 11 } \\ 2 & = 2 \:\:\color{Cerulean}{✓} \end{aligned}\)	\(\begin{aligned} \frac { 2 \left( \color{Cerulean}{ 5} \right) } { \left( 3 \left(\color{Cerulean}{ 5} \right) \color{black}{+} 1 \right) } & = \frac { 1 } { \left( \left( \color{Cerulean}{5 } \right) - 5 \right) } - \frac { 4 \left( \left( \color{Cerulean}{ 5} \right) \color{black}{-} 1 \right) } { \left( 3 \left( \color{Cerulean}{5 } \right) \color{black}{+} 1 \right) \left( \left( \color{Cerulean}{ 5 } \right) - 5 \right) } \\ \frac{10}{16} & = \frac{1}{0} - \frac{16}{0} \end{aligned}\)
	\(\color{Cerulean}{Undefined}\)

Here \(5\) is an extraneous solution and is not included in the solution set. It is important to note that \(5\) is a restriction.

Answer:

The solution is \(−12\).

If this process produces a solution that happens to be a restriction, then disregard it as a solution.

Exercise \(\PageIndex{1}\)

Solve: \(\frac { 4 ( x - 3 ) } { 36 - x ^ { 2 } } = \frac { 1 } { 6 - x } + \frac { 2 x } { 6 + x }\).

Answer

\(−\frac{3}{2}\)

www.youtube.com/v/Mxln5c53J1M

Sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. In the next two examples, we demonstrate two ways in which rational equation can have no solutions.

Example \(\PageIndex{4}\)

Solve: \(1 + \frac { 5 x + 22 } { x ^ { 2 } + 3 x - 4 } = \frac { x + 4 } { x - 1 }\)

Solution

To identify the LCD, first factor the denominators.

\(\begin{array} { c } { 1 + \frac { 5 x + 22 } { x ^ { 2 } + 3 x - 4 } = \frac { x + 4 } { x - 1 } } \\ { 1 + \frac { 5 x + 22 } { ( x + 4 ) ( x - 1 ) } = \frac { x + 4 } { ( x - 1 ) } } \end{array}\)

Multiply both sides by the LCD, \((x+4)(x−1)\), distributing carefully.

\(\begin{aligned} \color{Cerulean}{ ( x + 4 ) ( x - 1 )}\color{black}{ \cdot} \left( 1 + \frac { 5 x + 22 } { ( x + 4 ) ( x - 1 ) } \right) &= \color{Cerulean}{( x + 4 ) ( x - 1 )}\color{black}{ \cdot} \frac { x + 4 } { ( x - 1 ) } \\ \color{Cerulean}{( x + 4 ) ( x - 1 )}\color{black}{ \cdot} 1 +\color{Cerulean}{ ( x + 4 ) ( x - 1 )}\color{black}{ \cdot} \frac { ( 5 x + 22 ) } { ( x + 4 ) ( x - 1 ) }& = \color{Cerulean}{( x + 4 ) ( x - 1 ) }\color{black}{\cdot} \frac { ( x + 4 ) } { ( x - 1 ) } \\ (x+4)(x-1) + (5x+22) &=(x+4)(x+4) \\ x ^ { 2 } - x + 4 x - 4 + 5 x + 22& = x ^ { 2 } + 4 x + 4x + 16 \\ x ^ { 2 } + 8 x + 18 &= x ^ { 2 } + 8 x + 16 \\ 18 &= 16 \:\: \color{red} { False } \end{aligned}\)

The equation is a contradiction and thus has no solution.

Answer:

No solution, \(Ø\)

Example \(\PageIndex{5}\):

Solve: \(\frac { 3 x } { 2 x - 3 } - \frac { 3 ( 4 x + 3 ) } { 4 x ^ { 2 } - 9 } = \frac { x } { 2 x + 3 }\).

Solution

First, factor the denominators.

\(\frac { 3 x } { ( 2 x - 3 ) } - \frac { 3 ( 4 x + 3 ) } { ( 2 x + 3 ) ( 2 x - 3 ) } = \frac { x } { ( 2 x + 3 ) }\)

Take note that the restrictions on the domain are \(x≠±\frac{3}{2}\). To clear the fractions, multiply by the LCD, \((2x+3)(2x−3)\).

\(\begin{aligned} \frac { 3 x \cdot \color{Cerulean}{( 2 x + 3 ) ( 2 x - 3 )} } { \color{black}{( 2 x - 3 )} } - \frac { 3 ( 4 x + 3 ) \cdot \color{Cerulean}{( 2 x + 3 ) ( 2 x - 3 )} } {\color{black}{ (} 2 x + 3 ) ( 2 x - 3 ) } &= \frac { x \cdot \color{Cerulean}{ ( 2 x + 3 ) ( 2 x - 3 )} } { \color{black}{(} 2 x + 3 ) } \\ 3 x ( 2 x + 3 ) - 3 ( 4 x + 3 ) &= x ( 2 x - 3 ) \\ 6 x ^ { 2 } + 9 x - 12 x - 9 &= 2 x ^ { 2 } - 3 x \\ 6 x ^ { 2 } - 3 x - 9 &= 2 x ^ { 2 } - 3 x \\ 4 x ^ { 2 } - 9 &= 0 \\ ( 2 x + 3 ) ( 2 x - 3 ) &= 0 \end{aligned}\)

\(\begin{aligned} 2 x + 3 & = 0 \quad\quad \text { or }& 2 x - 3& = 0 \\ 2 x & = - 3 &2 x& = 3 \\ x &= - \frac { 3 } { 2 } & x& = \frac { 3 } { 2 } \end{aligned}\)

Both of these values are restrictions of the original equation; hence both are extraneous.

Answer:

No solution, \(Ø\)

It is important to point out that this technique for clearing algebraic fractions only works for equations. Do not try to clear algebraic fractions when simplifying expressions. As a reminder, an example of each is provided below.

Table \(\PageIndex{3}\)

Expression	Equation
\(\frac { 1 } { x } + \frac { x } { 2 x + 1 }\)	\(\frac { 1 } { x } + \frac { x } { 2 x + 1 } =0\)

Expressions are to be simplified and equations are to be solved. If we multiply the expression by the LCD, \(x (2x + 1)\), we obtain another expression that is not equivalent.

Table \(\PageIndex{4}\)

Incorrect	Correct
\(\begin{array} { l } { \frac { 1 } { x } + \frac { x } { 2 x + 1 } } \\ { \neq \color{red}{x ( 2 x + 1 )}\color{black}{ \cdot} \left( \frac { 1 } { x } + \frac { x } { 2 x + 1 } \right) } \\ { = 2 x + 1 + x ^ { 2 } } \color{red}{✗} \end{array}\)	\(\begin{aligned} \frac { 1 } { x } + \frac { x } { 2 x + 1 } & = 0 \\ \color{Cerulean}{x(2x+1)} \color{black}{\cdot} ( \frac { 1 } { x } + \frac { x } { 2 x + 1 } ) & = \color{Cerulean}{x ( 2 x + 1 )} \color{black}{\cdot}0 \\ 2 x + 1 + x ^ { 2 } & = 0 \\ x ^ { 2 } + 2 x + 1 & = 0 \color{Cerulean}{✓}\end{aligned}\)

Rational equations are sometimes expressed using negative exponents.

Example \(\PageIndex{6}\)

Solve: \(6+x^{−1}=x^{−2}\).

Solution^:

Begin by removing the negative exponents.

\(\begin{aligned} 6 + x ^ { - 1 } & = x ^ { - 2 } \\ 6 + \frac { 1 } { x } & = \frac { 1 } { x ^ { 2 } } \end{aligned}\)

Here we can see the restriction, \(x≠0\). Next, multiply both sides by the LCD, \(x^{2}\).

\(\begin{aligned} \color{Cerulean}{x ^ { 2 }}\color{black}{ \cdot} \left( 6 + \frac { 1 } { x } \right) & =\color{Cerulean}{ x ^ { 2} }\color{black}{ \cdot} \left( \frac { 1 } { x ^ { 2 } } \right) \\ \color{Cerulean}{x ^ { 2} }\color{black}{ \cdot} 6 + \color{Cerulean}{x ^ { 2} } \color{black}{\cdot} \frac { 1 } { x } & =\color{Cerulean}{ x ^ { 2} }\color{black}{ \cdot} \frac { 1 } { x ^ { 2 } } \\ 6 x ^ { 2 } + x & = 1 \\ 6 x ^ { 2 } + x - 1 & = 0 \\ ( 3 x - 1 ) ( 2 x + 1 ) & = 0 \end{aligned}\)

\(\begin{aligned} 3 x - 1 &= 0 \quad\quad \text { or } & 2 x + 1 &= 0 \\ 3 x &= 1 & 2 x & = - 1 \\ x &= \frac { 1 } { 3 } &x &= - \frac { 1 } { 2 } \end{aligned}\)

Answer:

\(−\frac{1}{2}, \frac{1}{3}\)

A proportion³⁵ is a statement of equality of two ratios.

\(\frac{a}{b}=\frac{c}{d}\)

This proportion is often read “\(a\) is to \(b\) as \(c\) is to \(d\).” Given any nonzero real numbers \(a, b, c\), and \(d\) that satisfy a proportion, multiply both sides by the product of the denominators to obtain the following:

\(\begin{aligned} \frac { a } { b } & = \frac { c } { d } \\ \color{Cerulean}{b d}\color{black}{ \cdot} \frac { a } { b } & = \color{Cerulean}{b d}\color{black}{ \cdot} \frac { c } { d } \\ a d & = b c \end{aligned}\)

This shows that cross products are equal, and is commonly referred to as cross multiplication³⁶.

If  \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{a}{d}=\frac{b}{c}\)

Cross multiply to solve proportions where terms are unknown.

Example \(\PageIndex{7}\)

Solve: \(\frac { 5 n - 1 } { 5 } = \frac { 3 n } { 2 }\).

Solution

When cross multiplying, be sure to group \(5n−1\).

Figure \(\PageIndex{1}\)

\(( 5 n - 1 ) \cdot 2 = 5 \cdot 3 n\)

Apply the distributive property in the next step.

\(\begin{aligned} ( 5 n - 1 ) \cdot 2 & = 5 \cdot 3 n \\ 10 n - 2 & = 15 n \quad \color{Cerulean} { Distribute. } \\ - 2 & = 5 n \quad\:\:\color{Cerulean}{Solve.} \\ \frac { - 2 } { 5 } & = n \end{aligned}\)

Answer:

\(n=−\frac{2}{5}\)

Cross multiplication can be used as an alternate method for solving rational equations. The idea is to simplify each side of the equation to a single algebraic fraction and then cross multiply.

Example \(\PageIndex{8}\)

Solve: \(\frac { 1 } { 2 } - \frac { 4 } { x } = - \frac { x } { 8 }\).

Solution

Obtain a single algebraic fraction on the left side by subtracting the equivalent fractions with a common denominator.

\(\begin{aligned} \frac { 1 } { 2 } \cdot \color{Cerulean}{\frac { x } { x }}\color{black}{ -} \frac { 4 } { x } \cdot \color{Cerulean}{\frac { 2 } { 2 }} & \color{black}{=} - \frac { x } { 8 } \\ \frac { x } { 2 x } - \frac { 8 } { 2 x } & = - \frac { x } { 8 } \\ \frac { x - 8 } { 2 x } & = - \frac { x } { 8 } \end{aligned}\)

Note that \(x≠0\), cross multiply, and then solve for \(x\).

\(\begin{aligned} \frac { x - 8 } { 2 x } & = \frac { - x } { 8 } \\ 8 ( x - 8 ) & = - x \cdot 2 x \\ 8 x - 64 & = - 2 x ^ { 2 } \\ 2 x ^ { 2 } + 8 x - 64 & = 0 \\ 2 \left( x ^ { 2 } + 4 x - 32 \right) & = 0 \\ 2 ( x - 4 ) ( x + 8 ) & = 0 \end{aligned}\)

Next, set each variable factor equal to zero.

\(\begin{aligned} x - 4 & = 0 \quad\quad\text { or } &x + 8 &= 0 \\ x & = 4 \quad &x &= - 8 \end{aligned}\)

The check is left to the reader.

Answer:

\(−8, 4\)

Exercise \(\PageIndex{2}\)

Solve: \(\frac { 2 ( 2 x - 5 ) } { x - 1 } = - \frac { x - 4 } { 2 x - 5 }\).

Answer

Answer: \(2, 3\)

www.youtube.com/v/KYzJwk_dxHc

Solving Literal Equations and Applications Involving Reciprocals

Literal equations, or formulas, are often rational equations. Hence the techniques described in this section can be used to solve for particular variables. Assume that all variable expressions in the denominator are nonzero.

Example \(\PageIndex{9}\):

The reciprocal of the combined resistance \(R\) of two resistors \(R_{1}\) and \(R_{2}\) in parallel is given by the formula \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\). Solve for \(R\) in terms of \(R_{1}\) and \(R_{2}\).

Figure \(\PageIndex{2}\)

Solution

The goal is to isolate \(R\) on one side of the equation. Begin by multiplying both sides of the equation by the LCD, \(RR_{1}R_{2}\).

\(\begin{aligned} \color{Cerulean}{R R _ { 1 } R _ { 2 }}\color{black}{ \cdot} \frac { 1 } { R } &= \color{Cerulean}{R R _ { 1 } R _ { 2} }\color{black}{ \cdot} \frac { 1 } { R _ { 1 } } +\color{Cerulean}{ R R _ { 1 } R _ { 2} }\color{black}{ \cdot} \frac { 1 } { R _ { 2 } } \\ R _ { 1 } R _ { 2 } & = R R _ { 2 } + R R _ { 1 } \\ R _ { 1 } R _ { 2 } &= R \left( R _ { 2 } + R _ { 1 } \right) \\ \frac { R _ { 1 } R _ { 2 } } { R _ { 2 } + R _ { 1 } } & = R \end{aligned}\)

Answer:

\(R = \frac { R _ { 1 } R _ { 2 } } { R _ { 1 } + R _ { 2 } }\)

Exercise \(\PageIndex{3}\)

Solve for \(y : x = \frac { 2 y + 5 } { y - 3 }\)

Answer

\(y = \frac { 3 x + 5 } { x - 2 }\)

www.youtube.com/v/YsSxYxGZBvA

Recall that the reciprocal of a nonzero number \(n\) is \(\frac{1}{n}\). For example, the reciprocal of \(5\) is \(\frac{1}{5}\) and \(5⋅\frac{1}{5}=1\). In this section, the applications will often involve the key word “reciprocal.” When this is the case, we will see that the algebraic setup results in a rational equation.

Example \(\PageIndex{10}\):

A positive integer is \(3\) less than another. If the reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger, then the result is \(\frac{1}{20}\). Find the two integers.

Solution

Let \(n\) represent the larger positive integer.

Let \(n − 3\) represent the smaller positive integer.

Set up an algebraic equation.

Figure \(\PageIndex{3}\)

Solve this rational expression by multiplying both sides by the LCD. The LCD is \(20n(n−3)\).

\(\begin{aligned} \frac { 2 } { n } - \frac { 1 } { n - 3 } &= \frac { 1 } { 20 } \\ \color{Cerulean}{20 n ( n - 3 )}\color{black}{ \cdot} \left( \frac { 2 } { n } - \frac { 1 } { n - 3 } \right) &= \color{Cerulean}{20 n ( n - 3 )}\color{black}{ \cdot} \left( \frac { 1 } { 20 } \right) \\ \color{Cerulean}{20 n ( n - 3 )}\color{black}{ \cdot} \frac { 2 } { n } - \color{Cerulean}{20 n ( n - 3 )}\color{black}{ \cdot} \frac { 1 } { n - 3 } &= \color{Cerulean}{20 n ( n - 3 )}\color{black}{ \cdot} \left( \frac { 1 } { 20 } \right) \\ 40(n-3) - 20n &= n(n-3) \\40n - 120 -20n &=n^{2}-3n\\20n-120&=n^{2}-3n \\0 &= n^{2}-23n+120\\0&=(n-8)(n-15) \end{aligned}\)

\(\begin{aligned} n - 8 & = 0 \quad\quad\text { or } & n - 15 &= 0 \\ n & = 8 \quad& n &= 15 \end{aligned}\)

Here we have two viable possibilities for the larger integer \(n\). For this reason, we will we have two solutions to this problem.

If \(n=8\), then \(n−3=8−3=5\).

If \(n=15\), then \(n−3=15−3=12\).

As a check, perform the operations indicated in the problem.

\(2 \left( \frac { 1 } { n } \right) - \frac { 1 } { n - 3 } = \frac { 1 } { 20 }\)

Check \(8\) and \(5\)	Check \(15\) and \(12\)
\(\begin{aligned} 2 \left( \frac { 1 } { \color{Cerulean}{8} } \right)\color{black}{ -} \frac { 1 } { \color{Cerulean}{5} } & \color{black}{=} \frac { 1 } { 4 } - \frac { 1 } { 5 } \\ & = \frac { 5 } { 20 } - \frac { 4 } { 20 } \\ & = \frac { 1 } { 20 } \:\:\color{Cerulean}{✓}\end{aligned}\)	\(\begin{aligned} 2 \left( \frac { 1 } { \color{Cerulean}{15} } \right) \color{black}{-} \frac { 1 } { \color{Cerulean}{12} } & \color{black}{=} \frac { 2 } { 15 } - \frac { 1 } { 12 } \\ & = \frac { 8 } { 60 } - \frac { 5 } { 60 } \\ & = \frac { 3 } { 60 } = \frac { 1 } { 20 } \:\:\color{Cerulean}{✓}\end{aligned}\)

Answer:

Two sets of positive integers solve this problem: \(\{ 5,8 \}\) and \(\{ 12,15 \}\).

Exercise \(\PageIndex{4}\)

When the reciprocal of the larger of two consecutive even integers is subtracted from \(4\) times the reciprocal of the smaller, the result is \(\frac{5}{6}\). Find the integers.

Answer

\(4, 6\)

www.youtube.com/v/z1t5VIN8yCI

Key Takeaways

• Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point.
• Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.
• When multiplying both sides of an equation by an expression, distribute carefully and multiply each term by that expression.
• If all of the resulting solutions are extraneous, then the original equation has no solutions.

Exercise \(\PageIndex{5}\)

Solve

1. \(\frac { 3 } { x } + 2 = \frac { 1 } { 3 x }\)
2. \(5 - \frac { 1 } { 2 x } = - \frac { 1 } { x }\)
3. \(\frac { 7 } { x ^ { 2 } } + \frac { 3 } { 2 x } = \frac { 1 } { x ^ { 2 } }\)
4. \(\frac { 4 } { 3 x ^ { 2 } } + \frac { 1 } { 2 x } = \frac { 1 } { 3 x ^ { 2 } }\)
5. \(\frac { 1 } { 6 } + \frac { 2 } { 3 x } = \frac { 7 } { 2 x ^ { 2 } }\)
6. \(\frac { 1 } { 12 } - \frac { 1 } { 3 x } = \frac { 1 } { x ^ { 2 } }\)
7. \(2 + \frac { 3 } { x } + \frac { 7 } { x ( x - 3 ) } = 0\)
8. \(\frac { 20 } { x } - \frac { x + 44 } { x ( x + 2 ) } = 3\)
9. \(\frac { 2 x } { 2 x - 3 } + \frac { 4 } { x } = \frac { x - 18 } { x ( 2 x - 3 ) }\)
10. \(\frac { 2 x } { x - 5 } + \frac { 2 ( 4 x + 7 ) } { x ( x - 5 ) } = - \frac { 1 } { x }\)
11. \(\frac { 4 } { 4 x - 1 } - \frac { 1 } { x - 1 } = \frac { 2 } { 4 x - 1 }\)
12. \(\frac { 5 } { 2 x - 3 } - \frac { 1 } { x + 3 } = \frac { 2 } { 2 x - 3 }\)
13. \(\frac { 4 x } { x - 3 } + \frac { 4 } { x ^ { 2 } - 2 x - 3 } = - \frac { 1 } { x + 1 }\)
14. \(\frac { 2 x } { x - 2 } - \frac { 15 } { x + 4 } = \frac { 24 } { x ^ { 2 } + 2 x - 8 }\)
15. \(\frac { x } { x - 8 } - \frac { 8 } { x - 1 } = \frac { 56 } { x ^ { 2 } - 9 x + 8 }\)
16. \(\frac { 2 x } { x - 1 } + \frac { 9 } { 3 x - 1 } + \frac { 11 } { 3 x ^ { 2 } - 4 x + 1 } = 0\)
17. \(\frac { 3 x } { x - 2 } - \frac { 14 } { 2 x ^ { 2 } - x - 6 } = \frac { 2 } { 2 x + 3 }\)
18. \(\frac { x } { x - 4 } - \frac { 4 } { x - 5 } = - \frac { 4 } { x ^ { 2 } - 9 x + 20 }\)
19. \(\frac { 2 x } { 5 + x } - \frac { 1 } { 5 - x } = \frac { 2 x } { x ^ { 2 } - 25 }\)
20. \(\frac { 2 x } { 2 x + 3 } - \frac { 1 } { 2 x - 3 } = \frac { 6 } { 9 - 4 x ^ { 2 } }\)
21. \(1 + \frac { 1 } { x + 1 } = \frac { 8 } { x - 1 } - \frac { 16 } { x ^ { 2 } - 1 }\)
22. \(1 - \frac { 1 } { 3 x + 5 } = \frac { 2 x } { 3 x - 5 } - \frac { 2 ( 6 x + 5 ) } { 9 x ^ { 2 } - 25 }\)
23. \(\frac { x } { x - 2 } - \frac { 3 } { x + 8 } = \frac { x + 2 } { x + 8 } + \frac { 5 ( x + 3 ) } { x ^ { 2 } + 6 x - 16 }\)
24. \(\frac { 2 x } { x - 10 } + \frac { 1 } { x - 3 } = \frac { x + 3 } { x - 10 } + \frac { x ^ { 2 } - 5 x + 5 } { x ^ { 2 } - 13 x + 30 }\)
25. \(\frac { 5 } { x ^ { 2 } + 9 x + 18 } + \frac { x + 3 } { x ^ { 2 } + 7 x + 6 } = \frac { 5 } { x ^ { 2 } + 4 x + 3 }\)
26. \(\frac { 1 } { x ^ { 2 } + 4 x - 60 } + \frac { x - 6 } { x ^ { 2 } + 16 x + 60 } = \frac { 1 } { x ^ { 2 } - 36 }\)
27. \(\frac { 4 } { x ^ { 2 } + 10 x + 21 } + \frac { 2 ( x + 3 ) } { x ^ { 2 } + 6 x - 7 } = \frac { x + 7 } { x ^ { 2 } + 2 x - 3 }\)
28. \(\frac { x - 1 } { x ^ { 2 } - 11 x + 28 } + \frac { x - 1 } { x ^ { 2 } - 5 x + 4 } = \frac { x - 4 } { x ^ { 2 } - 8 x + 7 }\)
29. \(\frac { 5 } { x ^ { 2 } + 5 x + 4 } + \frac { x + 1 } { x ^ { 2 } + 3 x - 4 } = \frac { 5 } { x ^ { 2 } - 1 }\)
30. \(\frac { 1 } { x ^ { 2 } - 2 x - 63 } + \frac { x - 9 } { x ^ { 2 } + 10 x + 21 } = \frac { 1 } { x ^ { 2 } - 6 x - 27 }\)
31. \(\frac { 4 } { x ^ { 2 } - 4 } + \frac { 2 ( x - 2 ) } { x ^ { 2 } - 4 x - 12 } = \frac { x + 2 } { x ^ { 2 } - 8 x + 12 }\)
32. \(\frac { x + 2 } { x ^ { 2 } - 5 x + 4 } + \frac { x + 2 } { x ^ { 2 } + x - 2 } = \frac { x - 1 } { x ^ { 2 } - 2 x - 8 }\)

Answer

1. \(−\frac{4}{3}\)

3. \(−4\)

5. \(−7, 3\)

7. \(−\frac{1}{2} , 2\)

9. \(−2, −\frac{3}{2}\)

11. \(−\frac{1}{2}\)

13. \(−\frac{1}{4}\)

15. \(Ø\)

17. \(−2, \frac{5}{6}\)

19. \(\frac{1}{2}\)

21. \(6\)

23. \(Ø\)

25. \(−8, 2\)

27. \(5\)

29. \(−6, 4\)

31. \(10\)

Exercise \(\PageIndex{6}\)

Solve the following equations involving negative exponents.

1. \(2 x ^ { - 1 } = 2 x ^ { - 2 } - x ^ { - 1 }\)
2. \(3 + x ( x + 1 ) ^ { - 1 } = 2 ( x + 1 ) ^ { - 1 }\)
3. \(x ^ { - 2 } - 64 = 0\)
4. \(1 - 4 x ^ { - 2 } = 0\)
5. \(x - ( x + 2 ) ^ { - 1 } = - 2\)
6. \(2 x - 9 ( 2 x - 1 ) ^ { - 1 } = 1\)
7. \(2 x ^ { - 2 } + ( x - 12 ) ^ { - 1 } = 0\)
8. \(- 2 x ^ { - 2 } + 3 ( x + 4 ) ^ { - 1 } = 0\)

Answer

1. \(\frac{2}{3}\)

3. \(\pm \frac { 1 } { 8 }\)

5. \(- 3 , - 1\)

7. \(- 6,4\)

Exercise \(\PageIndex{7}\)

Solve by cross multiplying.

1. \(\frac { 5 } { n } = - \frac { 3 } { n - 2 }\)
2. \(\frac { 2 n - 1 } { 2 n } = - \frac { 1 } { 2 }\)
3. \(- 3 = \frac { 5 n + 2 } { 3 n }\)
4. \(\frac { n + 1 } { 2 n - 1 } = \frac { 1 } { 3 }\)
5. \(\frac { x + 2 } { x - 5 } = \frac { x + 4 } { x - 2 }\)
6. \(\frac { x + 1 } { x + 5 } = \frac { x - 5 } { x }\)
7. \(\frac { 2 x + 1 } { 6 x - 1 } = \frac { x + 5 } { 3 x - 2 }\)
8. \(\frac { 6 ( 2 x + 3 ) } { 4 x - 1 } = \frac { 3 x } { x + 2 }\)
9. \(\frac { 3 ( x + 1 ) } { 1 - x } = \frac { x + 3 } { x + 1 }\)
10. \(\frac { 8 ( x - 2 ) } { x + 1 } = \frac { 5 - x } { x - 2 }\)
11. \(\frac { x + 3 } { x + 7 } = \frac { x + 3 } { 3 ( 5 - x ) }\)
12. \(\frac { x + 1 } { x + 4 } = \frac { - 8 ( x + 4 ) } { x + 7 }\)

Answer

1. \(\frac{5}{4}\)

3. \(-\frac{1}{7}\)

5. \(-16\)

7. \(\frac{1}{10}\)

9. \(-2,0\)

11. \(-3,2\)

Exercise \(\PageIndex{8}\)

Simplify or solve, whichever is appropriate.

1. \(\frac { 1 } { x } + \frac { 2 } { x - 3 } = - \frac { 2 } { 3 }\)
2. \(\frac { 1 } { x - 3 } - \frac { 3 } { 4 } = \frac { 1 } { x }\)
3. \(\frac { x - 2 } { 3 x - 1 } - \frac { 2 - x } { x }\)
4. \(\frac { 5 } { 2 } + \frac { x } { 2 x - 1 } - \frac { 1 } { 2 x }\)
5. \(\frac { x - 1 } { 3 x } + \frac { 2 } { x + 1 } - \frac { 5 } { 6 }\)
6. \(\frac { x - 1 } { 3 x } + \frac { 2 } { x + 1 } = \frac { 5 } { 6 }\)
7. \(\frac { 2 x + 1 } { 2 x - 3 } + 2 = \frac { 1 } { 2 x }\)
8. \(5 - \frac { 3 x + 1 } { 2 x } + \frac { 1 } { x + 1 }\)

Answer

1. Solve; \(- 3 , \frac { 3 } { 2 }\)

3. Simplify; \(\frac { ( 4 x - 1 ) ( x - 2 ) } { x ( 3 x - 1 ) }\)

5. Simplify; \(- \frac { ( x - 2 ) ( 3 x - 1 ) } { 6 x ( x + 1 ) }\)

7. Solve; \(\frac{1}{2}\)

Exercise \(\PageIndex{9}\)

Find the roots of the given function.

1. \(f ( x ) = \frac { 2 x - 1 } { x - 1 }\)
2. \(f ( x ) = \frac { 3 x + 1 } { x + 2 }\)
3. \(g ( x ) = \frac { x ^ { 2 } - 81 } { x ^ { 2 } - 5 x }\)
4. \(g ( x ) = \frac { x ^ { 2 } - x - 20 } { x ^ { 2 } - 9 }\)
5. \(f ( x ) = \frac { 4 x ^ { 2 } - 9 } { 2 x - 3 }\)
6. \(f ( x ) = \frac { 3 x ^ { 2 } - 2 x - 1 } { x ^ { 2 } - 1 }\)
7. Given \(f ( x ) = \frac { 1 } { x } + 5\), find \(x\) when \(f(x)=2\).
8. Given \(f ( x ) = \frac { 1 } { x - 4 }\), find \(x\) when \(f ( x ) = \frac { 1 } { 2 }\).
9. Given \(f ( x ) = \frac { 1 } { x + 3 } + 2\), find \(x\) when \(f(x)=1\).
10. Given \(f ( x ) = \frac { 1 } { x - 2 } + 5\), find \(x\) when \(f(x)=3\).

Answer

1. \(\frac{1}{2}\)

3. \(\pm 9\)

5. \(-\frac{3}{2}\)

7. \(x=-\frac{1}{3}\)

9. \(x=-4\)

Exercise \(\PageIndex{10}\)

Find the \(x\)- and \(y\)-intercepts.

1. \(f ( x ) = \frac { 1 } { x + 1 } + 4\)
2. \(f ( x ) = \frac { 1 } { x - 2 } - 6\)
3. \(f ( x ) = \frac { 1 } { x - 3 } + 2\)
4. \(f ( x ) = \frac { 1 } { x + 1 } - 1\)
5. \(f ( x ) = \frac { 1 } { x } - 3\)
6. \(f ( x ) = \frac { 1 } { x + 5 }\)

Answer

1. x-intercept: \((−\frac{5}{4} , 0)\); y-intercept: \((0, 5)\)

3. x-intercept: \((\frac{5}{2} , 0)\); y-intercept: \((0, \frac{5}{3})\)

5. x-intercept: \((\frac{1}{3} , 0)\); y-intercept: none

Exercise \(\PageIndex{11}\)

Find the points where the given functions coincide. (Hint: Find the points where \(f ( x ) = g ( x )\).)

1. \(f ( x ) = \frac { 1 } { x } , g ( x ) = x\)
2. \(f ( x ) = - \frac { 1 } { x } , g ( x ) = - x\)
3. \(f ( x ) = \frac { 1 } { x - 2 } + 3 , g ( x ) = x + 1\)
4. \(f ( x ) = \frac { 1 } { x + 3 } - 1 , g ( x ) = x + 2\)

Answer

1. \((−1, −1)\) and \((1, 1)\)

3. \((1, 2)\) and \((3, 4)\)

Exercise \(\PageIndex{12}\)

Recall that if \(| X | = p\), then \(X = - p\) or \(X=p\). Use this to solve the following absolute value equations.

1. \(\left| \frac { 1 } { x + 1 } \right| = 2\)
2. \(\left| \frac { 2 x } { x + 2 } \right| = 1\)
3. \(\left| \frac { 3 x - 2 } { x - 3 } \right| = 4\)
4. \(\left| \frac { 5 x - 3 } { 2 x + 1 } \right| = 3\)
5. \(\left| \frac { x ^ { 2 } } { 5 x + 6 } \right| = 1\)
6. \(\left| \frac { x ^ { 2 } - 48 } { x } \right| = 2\)

Answer

1. \(−\frac{3}{2} , −\frac{1}{2}\)

3. \(2, 10\)

5. \(−3, −2, −1, 6\)

Exercise \(\PageIndex{13}\)

Solve for the given variable.

1. Solve for \(P : w = \frac { P - 2 l } { 2 }\)
2. Solve for \(A : t = \frac { A - P } { P r }\)
3. Solve for \(t : \frac { 1 } { t _ { 1 } } + \frac { 1 } { t _ { 2 } } = \frac { 1 } { t }\)
4. Solve for \(n : P = 1 + \frac { r } { n }\)
5. Solve for \(y : m = \frac { y - y _ { 0 } } { x - x _ { 0 } }\)
6. Solve for \(m _ { 1 } : F = G \frac { m _ { 1 } m _ { 2 } } { r ^ { 2 } }\)
7. Solve for \(y : x = \frac { 2 y - 1 } { y - 1 }\)
8. Solve for \(y : x = \frac { 3 y + 2 } { y + 3 }\)
9. Solve for \(y : x = \frac { 2 y } { 2 y + 5 }\)
10. Solve for \(y : x = \frac { 5 y + 1 } { 3 y }\)
11. Solve for \(x : \frac { a } { x } + \frac { c } { b } = \frac { a } { c }\)
12. Solve for \(y : \frac { a } { y } - \frac { 1 } { a } = b\)

Answer

1. \(P = 2 l + 2 w\)

3. \(t = \frac { t _ { 1 } t _ { 2 } } { t _ { 1 } + t _ { 2 } }\)

5. \(y = m \left( x - x _ { 0 } \right) + y _ { 0 }\)

7. \(y = \frac { x - 1 } { x - 2 }\)

9. \(y = - \frac { 5 x } { 2 x - 2 }\)

11. \(x = \frac { a b c } { a b - c ^ { 2 } }\)

Exercise \(\PageIndex{14}\)

Use algebra to solve the following applications.

1. The value in dollars of a tablet computer is given by the function \(V ( t ) = 460 ( t + 1 ) ^ { - 1 }\), where \(t\) represents the age of the tablet. Determine the age of the tablet if it is now worth \($100\).
2. The value in dollars of a car is given by the function \(V ( t ) = 24,000 ( 0.5 t + 1 ) ^ { - 1 }\), where \(t\) represents the age of the car. Determine the age of the car if it is now worth \($6,000\).

Answer

1. \(3.6\) years old

Exercise \(\PageIndex{15}\)

Solve for the unknowns.

1. When \(2\) is added to \(5\) times the reciprocal of a number, the result is \(12\). Find the number.
2. When \(1\) is subtracted from \(4\) times the reciprocal of a number, the result is \(11\). Find the number.
3. The sum of the reciprocals of two consecutive odd integers is \(\frac{12}{35}\). Find the integers.
4. The sum of the reciprocals of two consecutive even integers is \(\frac{9}{40}\). Find the integers.
5. An integer is \(4\) more than another. If \(2\) times the reciprocal of the larger is subtracted from \(3\) times the reciprocal of the smaller, then the result is \(\frac{1}{8}\). Find the integers.
6. An integer is \(2\) more than twice another. If \(2\) times the reciprocal of the larger is subtracted from \(3\) times the reciprocal of the smaller, then the result is \(\frac{5}{14}\). Find the integers.
7. If \(3\) times the reciprocal of the larger of two consecutive integers is subtracted from \(2\) times the reciprocal of the smaller, then the result is \(\frac{1}{2}\). Find the two integers.
8. If \(3\) times the reciprocal of the smaller of two consecutive integers is subtracted from \(7\) times the reciprocal of the larger, then the result is \(\frac{1}{2}\). Find the two integers.
9. A positive integer is \(5\) less than another. If the reciprocal of the smaller integer is subtracted from \(3\) times the reciprocal of the larger, then the result is \(\frac{1}{12}\). Find the two integers.
10. A positive integer is \(6\) less than another. If the reciprocal of the smaller integer is subtracted from \(10\) times the reciprocal of the larger, then the result is \(\frac{3}{7}\). Find the two integers.

Answer

1. \(\frac{1}{2}\)

3. \(5,7\)

5. \(\{ - 8 , - 4 \} \text { and } \{ 12,16 \}\)

7. \(\{ 1,2 \} \text { or } \{ - 4 , - 3 \}\)

9. \(\{ 4,9 \} \text { or } \{ 15,20 \}\)

Exercise \(\PageIndex{16}\)

1. Explain how we can tell the difference between a rational expression and a rational equation. How do we treat them differently? Give an example of each.
2. Research and discuss reasons why multiplying both sides of a rational equation by the LCD sometimes produces extraneous solutions.

Answer

1. Answer may vary

Footnotes

³³An equation containing at least one rational expression.

³⁴A solution that does not solve the original equation.

³⁵A statement of equality of two ratios.

³⁶If \(\frac{a}{b} = \frac{c}{d}\) then \(ad = bc\).