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# Solving Rational Equations using Cross-Multiplication

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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 $2x-21$ units, and the length of the real racecar is $x^2$ units. What is the value of x ?

### Guidance

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 in the Geometry FlexBook. Otherwise, we will start solving rational equations using cross-multiplication.

#### Example A

Solve $\frac{x}{2x-3}=\frac{3x}{x+11}$ .

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.

$\frac{0}{2 \cdot 0-3}&=\frac{3 \cdot 0}{0+11} && \frac{4}{2 \cdot 4-3}=\frac{3 \cdot 4}{4+11} \\\frac{0}{-3}&=\frac{0}{11} \ && \qquad \quad \frac{4}{5}=\frac{12}{15} \\0&=0 && \qquad \quad \frac{4}{5}=\frac{4}{5}$

#### Example B

Solve $\frac{x+1}{4}=\frac{3}{x-3}$ .

Solution: Cross-multiply and solve.

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

$\frac{5+1}{4}=\frac{3}{5-3} \rightarrow \frac{6}{4}=\frac{3}{2}$ and

$\frac{-3+1}{4}=\frac{3}{-3-3} \rightarrow \frac{-2}{4}=\frac{3}{-6}$

#### Example C

Solve $\frac{x^2}{2x-5}=\frac{x+8}{2}$ .

Solution: Cross-multiply.

$\frac{x^2}{2x-5}&=\frac{x+8}{2} \\2x^2+11x-40&=2x^2 \\11x-40&=0 \\11x&=40 \\x&=\frac{40}{11}$

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

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

$\frac{1}{x} = \frac{2x-21}{x^2}$

Now cross-multiply.

$x^2 = x(2x-21)\\x^2 = 2x^2 - 21x\\0 = x^2 - 21x\\0 = x(x - 21)\\x = 0, 21$

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. $\frac{-x}{x-1}=\frac{x-8}{3}$

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

3. $\frac{9-x}{x^2}=\frac{4}{3x}$

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

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

2. $\frac{x^2-1}{x+2}&=\frac{2x-1}{2} \\2x^2+3x-2&=2x^2-2\\3x&=0 \\x&=0$

$\text{Check}: \frac{0^2-1}{0+2}&=\frac{2 \left(0\right)-1}{2} \\\frac{-1}{2}&=\frac{-1}{2}$

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

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

$x = 0$ 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.

### Vocabulary

Rational Equation
An equation where there are rational expressions on both sides of the equal sign.

### Problem Set

1. Is $x=-2$ a solution to $\frac{x-1}{x-4}=\frac{x^2-1}{x+4}$ ?

Solve the following rational equations.

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

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

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