Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 2 - Section 2.3 - Solving Linear Equations - Exercise Set - Page 142: 81


The expression is evaluated to $240$.

Work Step by Step

We first need to solve the equation to find out what $x$ is. We will multiply the entire equation by the least common denominator in order to get rid of the fractions. The least common denominator is $15$, so we multiply the entire equation by $15$: $$15(\frac{x}{5}) - 15(2) = 15(\frac{x}{3})$$ Divide out common factors to get rid of the fractions: $$3x - 30 = 5x$$ Subtract $3x$ from each side to get: $$-30 = 2x$$ Solve for $x$: $$x = -15$$ Now that we have the value for $x$, we can plug it into the expression $x^{2} - x$: $$(-15)^{2} - (-15)$$ We square $-15$ first to get: $$225 - (-15)$$ A negative and a negative make a positive, so we can rewrite the expression as follows: $$225 + 15$$ The expression evaluates to $$240$$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.