Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 5 - Linear Functions - 5-5 Standard Form - Practice and Problem-Solving Exercises - Page 328: 67



Work Step by Step

First, we need to move all constants to the right side of the equation. To do this, we add $19$ to each side of the equation: $\frac{7}{2}x = -13 + 19 + 2x$ Next, we want to move all $x$ terms to the left side of the equation. To do this, we subtract $2x$ from both sides of the equation: $\frac{7}{2}x - 2x = -13 + 19$ Let's change $2x$ to a fraction with $2$ as its denominator so we can add the two fractions: $\frac{7}{2}x - \frac{4}{2}x = -13 + 19$ Combine like terms: $\frac{3}{2}x = 6$ To solve for $x$, we divide both sides by $\frac{3}{2}$. This means that we multiply by the reciprocal of this fraction, which is $\frac{2}{3}$: $x = 6(\frac{2}{3})$ Multiply to simplify: $x = \frac{12}{3}$ Simplify the fraction by dividing both the numerator and denominator by their greatest common factor, which is $3$: $x = 4$ This answer corresponds to option $H$.
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