## Solving a Rational Equation

This post about solving a rational equation is #15 in a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Solve this rational equation:

$\fn_jvn {\color{Blue} \frac{5}{x+8}=\frac{x}{3x-2}}$

Assume x is not equal to -8 or 2/3.

## Solution

What do we mean by rational? The root is ratio. A rational number is a number that can be expressed as quotient — a ratio — of two integers. A rational expression is a quotient — a ratio — of algebraic expressions. A rational equation is an equation comprised of rational expressions.

To get started solving this rational equation, multiply both sides by $\inline \fn_jvn \left (x+8 \right )\left (3x+2 \right )$. That should give you:

$\fn_jvn {\color{Blue} \frac{5\cdot \left (x+8 \right )\left (3x+2 \right )}{x+8}=\frac{x\left (x+8 \right )\left (3x+2 \right )}{3x-2}}$

Cancel the terms that show up in both numerator and denominator. That’s $\inline \fn_jvn x+8$ on the left and $\inline \fn_jvn 3x-2$ on the right. That leaves:

$\fn_jvn {\color{Blue} 5 \cdot \left (3x-2 \right )=x\left (x+8 \right )}$

(Yes, I could have just said “cross multiply,” but I hate that expression. It’s a personal quirk.) Now we’re done with denominators.

Distribute:

$\fn_jvn {\color{Blue} 15x-10=x^2+8x}$

Subtract $\inline \fn_jvn 15x-10$ from both sides, and then switch the sides.

$\fn_jvn {\color{Blue} 0=x^2-7x+10}$

$\fn_jvn {\color{Blue} x^2-7x+10=0}$

Factor:

$\fn_jvn {\color{Blue} x^2-7x+10=\left (x-2 \right )\left ( x-5 \right )}$

Set each factor equal to 0:

 $\inline \fn_jvn x-2=0$ $\inline \fn_jvn x-5=0$ $\inline \fn_jvn x=2$ $\inline \fn_jvn x=5$

### Caution

Remember that division by zero yields something undefined, so an x-value that makes a denominator zero is not a valid answer. So you need to make sure that the x values you’ve found do not make one of the denominators in the original question zero. To find those values, set those denominators equal to zero:

 $\inline \fn_jvn x+8=0$ $\inline \fn_jvn 3x-2=0$ $\inline \fn_jvn x=-8$ $\inline \fn_jvn x=2/3$

Oh, look: Those are the values that the question says are excluded. The question is telling you that a solution that causes the denominator to be zero is not a valid solution. But today that is not your problem, because that’s not one of the answers we found. We found $\inline \fn_jvn x=2$ and $\inline \fn_jvn x=5$. Interesting factoid: The official practice question on which this question is based stipulates that x is not equal to -3/2 and then offers -3/2 as an answer choice. If there’s a lesson in that, it may be that it’s a good idea to reread the question before you give your final answer.

To complete the check, substitute $\inline \fn_jvn x=2$ and $\inline \fn_jvn x=5$ into the original equation.

This question is similar to question number 15 in the sample questions for the Accuplacer Advanced Algebra and Functions test.