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Solving Rational Equations using Cross-Multiplication

Solve equations that are fractions on both sides

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Solving Rational Equations using Cross-Multiplication

A scale model of a racecar is in the ratio of 1:x to the real racecar. The length of the model is \begin{align*}2x-21\end{align*} units, and the length of the real racecar is \begin{align*}x^2\end{align*} units. What is the value of x?


A rational equation is an equation where there are rational expressions on both sides of the equal sign. One way to solve rational equations is to use cross-multiplication. Here is an example of a proportion that we can solve using cross-multiplication.

If you need more of a review of cross-multiplication, see the Proportion Properties concept. Otherwise, we will start solving rational equations using cross-multiplication.

Example A

Solve \begin{align*}\frac{x}{2x-3}=\frac{3x}{x+11}\end{align*}.

Solution: Use cross-multiplication to solve the problem. You can use the example above as a guideline.

Check your answers. It is possible to get extraneous solutions with rational expressions.

\begin{align*}\frac{0}{2 \cdot 0-3}&=\frac{3 \cdot 0}{0+11} && \frac{4}{2 \cdot 4-3}=\frac{3 \cdot 4}{4+11} \\ \frac{0}{-3}&=\frac{0}{11} \ && \qquad \quad \frac{4}{5}=\frac{12}{15} \\ 0&=0 && \qquad \quad \frac{4}{5}=\frac{4}{5}\end{align*}

Example B

Solve \begin{align*}\frac{x+1}{4}=\frac{3}{x-3}\end{align*}.

Solution: Cross-multiply and solve.

\begin{align*}\frac{x+1}{4}&=\frac{3}{x-3} \\ 12&=x^2-2x-3 \\ 0&=x^2-2x-15 \\ 0&=(x-5)(x+3) \\ x&=5 \ and \ -3\end{align*}

Check your answers.

\begin{align*}\frac{5+1}{4}=\frac{3}{5-3} \rightarrow \frac{6}{4}=\frac{3}{2}\end{align*} and

\begin{align*}\frac{-3+1}{4}=\frac{3}{-3-3} \rightarrow \frac{-2}{4}=\frac{3}{-6} \end{align*}

Example C

Solve \begin{align*}\frac{x^2}{2x-5}=\frac{x+8}{2}\end{align*}.

Solution: Cross-multiply.

\begin{align*}\frac{x^2}{2x-5}&=\frac{x+8}{2} \\ 2x^2+11x-40&=2x^2 \\ 11x-40&=0 \\ 11x&=40 \\ x&=\frac{40}{11}\end{align*}

Check the answer: \begin{align*}\frac{\left(\frac{40}{11}\right)^2}{\frac{80}{11}-5}=\frac{\frac{40}{11}+8}{2} \rightarrow \frac{1600}{121} \div \frac{25}{11}=\frac{128}{11} \div 2 \rightarrow \frac{64}{11}=\frac{128}{22}\end{align*}

Intro Problem Revisit We need to set up a rational equation and solve for x.

\begin{align*}\frac{1}{x} = \frac{2x-21}{x^2}\end{align*}

Now cross-multiply.

\begin{align*}x^2 = x(2x-21)\\ x^2 = 2x^2 - 21x\\ 0 = x^2 - 21x\\ 0 = x(x - 21)\\ x = 0, 21\end{align*}

However, x is a ratio so it must be greater than 0. Therefore x equals 21 and the model is in the ratio 1:21 to the real racecar.

Guided Practice

Solve the following rational equations.

1. \begin{align*}\frac{-x}{x-1}=\frac{x-8}{3}\end{align*}

2. \begin{align*}\frac{x^2-1}{x+2}=\frac{2x-1}{2}\end{align*}

3. \begin{align*}\frac{9-x}{x^2}=\frac{4}{3x}\end{align*}


1. \begin{align*}\frac{-x}{x-1}&=\frac{x-8}{3} \\ x^2-9x+8&=-3x \\ x^2-6x+8&=0 \\ (x-4)(x-2)&=0\\ x&=4 \ and \ 2\end{align*}

\begin{align*}\text{Check}: x=4 \rightarrow \frac{-4}{4-1}&=\frac{4-8}{3} && x=2 \rightarrow \frac{-2}{2-1}=\frac{2-8}{3} \\ \frac{-4}{3}&=\frac{-4}{3} && \qquad \qquad \quad \frac{-2}{1}=\frac{-6}{3} \end{align*}

2. \begin{align*}\frac{x^2-1}{x+2}&=\frac{2x-1}{2} \\ 2x^2+3x-2&=2x^2-2\\ 3x&=0 \\ x&=0\end{align*}

\begin{align*}\text{Check}: \frac{0^2-1}{0+2}&=\frac{2 \left(0\right)-1}{2} \\ \frac{-1}{2}&=\frac{-1}{2}\end{align*}

3. \begin{align*}\frac{9-x}{x^2}&=\frac{4}{-3x} \\ 4x^2&=-27x+3x^2 \\ x^2+27x&=0 \\ x(x+27)&=0 \\ x&=0 \ and \ -27 \end{align*}

\begin{align*}\text{Check}: x=0 \rightarrow \frac{9-0}{0^2}&=\frac{4}{-3 \left(0\right)} && x=-27 \rightarrow \frac{9+27}{\left(-27\right)^2}=\frac{4}{-3 \left(-27\right)} \\ und&=und && \qquad \qquad \qquad \quad \frac{36}{729}=\frac{4}{81} \\ & && \qquad \qquad \qquad \quad \ \frac{4}{81}=\frac{4}{81}\end{align*}

\begin{align*}x = 0\end{align*} is not actually a solution because it is a vertical asymptote for each rational expression, if graphed. Because zero is not part of the domain, it cannot be a solution, and is extraneous.

Problem Set

  1. Is \begin{align*}x=-2\end{align*} a solution to \begin{align*}\frac{x-1}{x-4}=\frac{x^2-1}{x+4}\end{align*}?

Solve the following rational equations.

  1. \begin{align*}\frac{2x}{x+3}=\frac{8}{x}\end{align*}
  2. \begin{align*}\frac{4}{x+1}=\frac{x+2}{3}\end{align*}
  3. \begin{align*}\frac{x^2}{x+2}=\frac{x+3}{2}\end{align*}
  4. \begin{align*}\frac{3x}{2x-1}=\frac{2x+1}{x}\end{align*}
  5. \begin{align*}\frac{x+2}{x-3}=\frac{x}{3x-2}\end{align*}
  6. \begin{align*}\frac{x+3}{-3}=\frac{2x+6}{x-3}\end{align*}
  7. \begin{align*}\frac{2x+5}{x-1}=\frac{2}{x-4}\end{align*}
  8. \begin{align*}\frac{6x-1}{4x^2}=\frac{3}{2x+5}\end{align*}
  9. \begin{align*}\frac{5x^2+1}{10}=\frac{x^3-8}{2x}\end{align*}
  10. \begin{align*}\frac{x^2-4}{x+4}=\frac{2x-1}{3}\end{align*}

Determine the values of a that make each statement true. If there no values, write none.

  1. \begin{align*}\frac{1}{x-a}=\frac{x}{x+a}\end{align*}, such that there is no solution.
  2. \begin{align*}\frac{1}{x-a}=\frac{x}{x-a}\end{align*}, such that there is no solution.
  3. \begin{align*}\frac{x-a}{x}=\frac{1}{x+a}\end{align*}, such that there is one solution.
  4. \begin{align*}\frac{1}{x+a}=\frac{x}{x-a}\end{align*}, such that there are two integer solutions.

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 9.14. 


Rational Equation

Rational Equation

A rational equation is an equation that contains a rational expression.

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