7.4 Solve Rational Equations

Topics covered in this section are:

After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. 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 a rational equation.

RATIONAL EQUATION

rational equation is an equation that contains a rational expression.

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

 Rational Expression Rational Equation $\frac{1}{8}x+\frac{1}{2}$ $\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}$ $\frac{y+6}{y^{2}-36}$ $\frac{y+6}{y^{2}-36}=y+1$ $\frac{1}{n-3} + \frac{1}{n+4}$ $\frac{1}{n-3} + \frac{1}{n+4}=\frac{15}{n^{2}+n-12}$

7.4.1 Solve Rational Equations

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

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

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

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

EXTRANEOUS SOLUTION TO A RATIONAL EQUATION

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 $x≠c$ next to the equation.

Example 1

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

Solution

The steps of this method are shown.

HOW TO: Solve equations with rational expressions.

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

Example 2

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

Solution

 $1-\frac{5}{y}=-\frac{6}{y^{2}}$ Note any value of the variable that would make any denominator zero. $1-\frac{5}{y}=-\frac{6}{y^{2}}, y≠0$ Find the least common denominator of all denominators is the equation. The LCD is $y^{2}$. Clear the fractions by multiplying both sides of the equation by the LCD. $\textcolor{red}{y^{2}} \left(1-\frac{5}{y}\right)=\textcolor{red}{y^{2}} \left(-\frac{6}{y^{2}}\right)$ Distribute. $\textcolor{red}{y^{2}} \cdot 1-\textcolor{red}{y^{2}} \left(\frac{5}{y}\right)=y^{2} \left(-\frac{6}{y^{2}}\right)$ Multiply. $y^{2}-5y=-6$ Solve the resulting equation. First write the quadratic equation in standard form. $y^{2}-5y+6=0$ Factor. $(y-2)(y-3)=0$ Use the Zero Product Property. $y-2=0$ or $y-3=0$ Solve. $y=2$ or $y=3$ Check. We did not get $0$ as an algebraic solution. The solution is $y=2$, $y=3$.

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

Example 3

Solve: $\frac{2}{x+2}+\frac{4}{x-2}=\frac{x-1}{x^{2}-4}$.

Solution

 $\frac{2}{x+2}+\frac{4}{x-2}=\frac{x-1}{x^{2}-4}$ Note any value of the variable that would make any denominator zero. $\frac{2}{x+2}+\frac{4}{x-2}=\frac{x-1}{(x+2)(x-2)}, x≠-2, x≠2$ Find the least common denominator of all denominators is the equation. The LCD is $(x+2)(x-2)$. Clear the fractions by multiplying both sides of the equation by the LCD. $\textcolor{red}{(x+2)(x-2)} \left(\frac{2}{x+2}+\frac{4}{x-2}\right)=\textcolor{red}{(x+2)(x-2)} \left(\frac{x-1}{(x+2)(x-2)}\right)$ Distribute. $\textcolor{red}{(x+2)(x-2)} \frac{2}{x+2}+\textcolor{red}{(x+2)(x-2)}\frac{4}{x-2}$$=\textcolor{red}{(x+2)(x-2)} \left(\frac{x-1}{(x+2)(x-2)}\right) Remove common factors. \textcolor{red}{\cancel{(x+2)}(x-2)} \frac{2}{\cancel{x+2}}+\textcolor{red}{(x+2)\cancel{(x-2)}}\frac{4}{\cancel{x-2}}$$ =\textcolor{red}{\cancel{(x+2)(x-2)}} \left(\frac{x-1}{\cancel{(x+2)(x-2)}}\right)$ Simplify. $2(x-2)+4(x+2)=x-1$ Distribute. $2x-4+4x+8=x-1$ Solve. $6x+4=x-1$$5x=-5$$x=-1$ Check. We did not get $2$ or $-2$ as algebraic solutions. The solution is $x=-1$.

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

Example 4

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

Solution

 $\frac{m+11}{m^{2}-5m+4}=\frac{5}{m-4}-\frac{3}{m-1}$ Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator. $\frac{m+11}{(m-4)(m-1)}=\frac{5}{m-4}-\frac{3}{m-1}, m≠4, m≠1$ Find the least common denominator of all denominators is the equation. The LCD is $(m-4)(m-1)$. Clear the fractions by multiplying both sides of the equation by the LCD. $\textcolor{red}{(m-4)(m-1)} \left(\frac{m+11}{(m-4)(m-1}\right)$$=\textcolor{red}{(m-4)(m-1)} \left(\frac{5}{m-4}-\frac{3}{m-1}\right) Distribute. \textcolor{red}{(m-4)(m-1)} \left(\frac{m+11}{(m-4)(m-1}\right)$$=\textcolor{red}{(m-4)(m-1)} \frac{5}{m-4}-\textcolor{red}{(m-4)(m-1)} \frac{3}{m-1}$ Remove common factors. $\textcolor{red}{\cancel{(m-4)(m-1)}} \left(\frac{m+11}{\cancel{(m-4)(m-1}}\right)$$=\textcolor{red}{\cancel{(m-4)}(m-1)} \frac{5}{m-4}-\textcolor{red}{(m-4)\cancel{(m-1)}} \frac{3}{\cancel{m-1}} Simplify. m+11=5(m-1)-3(m-4) Distribute. m+11=5m-5-3m+12 Solve. m+11=2m+7$$4=m$ Check. The only algebraic solution was $4$, but we said that $4$ would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation.

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

Example 5

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

Solution

 $\frac{y}{y+6}=\frac{72}{y^{2}-36}+4$ Factor all the denominators, so we can note any value of the variable that would make any denominator zero. $\frac{y}{y+6}=\frac{72}{(y-6)(y+6)}+4, y≠6, y≠-6$ Find the least common denominator of all denominators is the equation. The LCD is $(y-6)(y+6)$. Clear the fractions. $\textcolor{red}{(y-6)(y+6)} \left(\frac{y}{y+6}\right)=\textcolor{red}{(y-6)(y+6)} \left(\frac{72}{(y-6)(y+6)}+4\right)$ Simplify. $(y-6) \cdot y=72+(y-6)(y+6)\cdot 4$ Simplify. $y(y-6)=72+4(y^{2}-36)$ Solve. $y^{2}-6y=72+4y^{2}-144$$0=3y^{2}+6y-72$$0=3(y^{2}+2y-24)$$0=3(y+6)(y-4)$$y=-6, y=4$ Check. The solution is $y=4$.

In some cases, all the algebraic expressions are extraneous.

Example 6

Solve: $\frac{x}{2x-2}-\frac{2}{3x+3}=\frac{5x^{2}-2x+9}{12x^{2}-12}$.

Solution

7.4.2 Use Rational Functions

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

Example 8

For a rational function, $f(x)=\frac{2x-6}{x^{2}-8x+15}$

• find the domain of the function
• solve $f(x)=1$
• find the points on the graph at this function value
Solution

Part 1

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

 $x^{2}-8x+15=0$ Factor the trinomial. $(x-3)(x-5)=0$ Use the Zero Product Property. $x-3=0 \ \ \ \ \ \ \ x-5=0$ Solve. $x=3 \ \ \ x=5$ The domain is all real numbers except $x≠3$, $x≠5$

Part 2

 $f(x)=1$ Substitute the expression. $\frac{2x-6}{x^{2}-8x+15}=1$ Factor the denominator. $\frac{2x-6}{(x-3)(x-5)}=1$ Multiply both sides by the LCD.$(x-3)(x-5)$. $\textcolor{red}{(x-3)(x-5)} \left(\frac{2x-6}{(x-3)(x-5)}\right)=\textcolor{red}{(x-3)(x-5)} (1)$ Simplify. $2x-6=x^{2}-8x+15$ Solve. $0=x^{2}-10x+21$ Factor. $0=(x-7)(x-3)$ Use the Zero Product Property. $x-7=0 \ \ \ \ \ \ x-3=0$ Solve. $x=7 \ \ \ \ \ x=3$

However, $x=3$ is outside the domain of this function, so we discard that root as extraneous.

Part 3

The root of the function is $1$ when $x=7$. So the points on the graph of this function when $f(x)=1$ is $(7,1)$.

7.4.3 Solve a Rational Equation for a Specific Variable

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

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

 $m=\frac{y-y_1}{x-x_1}$ Multiply both sides of the equation by $x-x_1$ $m(x-x_1)=(\frac{y-y_1}{x-x_1})(x-x_1)$ Simplify. $m(x-x_1)=y-y_1$ Rewrite the equation with the $y$ terms on the left. $y-y_1=m(x-x_1)$

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point $(2, 3)$. We will add one more step to solve for $y$.

Example 9

Solve $m=\frac{y-2}{x-3}$ for $y$.

Solution

 $m=\frac{\textcolor{red}{y}-2}{x-3}$ Note any value of the variable that would make the denominator zero. $m=\frac{\textcolor{red}{y}-2}{x-3}, x≠3$ Clear the fractions by multiplying both sides of the equation by the LCD, $x-3$. $(x-3)m=(x-3)\left(\frac{\textcolor{red}{y}-2}{x-3}\right)$ Simplify. $xm-3m=\textcolor{red}{y}-2$ Isolate the $y$ term. $xm-3m+2=\textcolor{red}{y}$

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

Example 10

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

Solution

 $\frac{1}{c}+\frac{1}{m}=1$ Note any value of the variable that would make the denominator zero. $\frac{1}{c}+\frac{1}{m}=1, c≠0, m≠0$ Clear the fractions by multiplying both sides of the equation by the LCD, $cm$. $\textcolor{red}{cm}\left(\frac{1}{c}+\frac{1}{m}\right)=\textcolor{red}{cm}(1)$ Distribute. $\textcolor{red}{cm}(\frac{1}{c})+\textcolor{red}{cm}(\frac{1}{m})=\textcolor{red}{cm}(1)$ Simplify. $m+c=cm$ Collect the terms with $c$ to the right. $m=cm-c$ Factor the expression on the right. $m=c(m-1)$ To isolate $c$, divide both sides by $m-1$. $\frac{m}{\textcolor{red}{m-1}}=\frac{c(m-1)}{\textcolor{red}{m-1}}$ Simplify by removing common factors. $\frac{m}{m-1}=c$

Notice that even though we excluded $c=0, m=0$ from the original equation, must must also now state that $m≠1$.