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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?

### Solving Rational Equations Using Cross Multiplication

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.

Let's use cross multiplication to solve the following equations.

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

Cross-multiply and solve.

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*}

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

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*}

\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*}

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

Cross-multiply and solve.

\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*}

### Examples

#### Example 1

Earlier, you were asked to find the value of x.

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.

Solve the following rational equations.

#### Example 2

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

\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*}

#### Example 3

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

\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*}

#### Example 4

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

\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.

### Review

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.

To see the Review answers, open this PDF file and look for section 9.14.

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### Vocabulary Language: English

Rational Equation

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