Algebra 1: Common Core (15th Edition)

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

Chapter 2 - Solving Equations - 2-1 Solving One-Step Equations - Practice and Problem-Solving Exercises - Page 86: 57

Answer

z=7$\frac{1}{3}$

Work Step by Step

In order to solve this algebraic expression, we must isolate the variable. A variable is the letter in the problem (for instance, in the equation x+2=10, x is the variable). Because addition and subtraction are inverse operations (they cancel each other out), we know we must add when a number is being subtracted from the variable and subtract when a number is being added to the variable. In the equation z-4$\frac{2}{3}$=2$\frac{2}{3}$, we must get rid of the 4$\frac{2}{3}$, which is on the same side as z, in order to solve. Thus, we add 4$\frac{2}{3}$ on both sides of the equation to find that z=7$\frac{1}{3}$. We plug 7$\frac{1}{3}$ in for z and confirm that our answer is, indeed, correct.
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