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: 82


The expression evaluates to $6$.

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 $4$, so we multiply the entire equation by $4$: $$4(\frac{3x}{2}) + 4(\frac{3x}{4}) = 4(\frac{x}{4}) - 4(4)$$ Divide out common factors to get rid of the fractions: $$6x + 3x = x - 16$$ Subtract $x$ from each side and combine like terms to get: $$8x = -16$$ Solve for $x$: $$x = -2$$ Now that we have the value for $x$, we can plug it into the expression $x^{2} - x$: $$(-2)^{2} - (-2)$$ We square $2$ first to get: $$4 - (-2)$$ A negative and a negative make a positive, so we can rewrite the equation as follows: $$4 + 2$$ The expression evaluates to $6$.
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