Algebra 1: Common Core (15th Edition)

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

Chapter 1 - Foundations for Algebra - 1-2 Order of Operations and Evaluating Expressions - Practice and Problem-Solving Exercises - Page 15: 59


The left side of the equation is equal to the right side of the equation, $28=28$, as proven by solving using PEMDAS.

Work Step by Step

$3\times(4+5)-6+7=28$ Use PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) to solve the equation correctly. 1. The first step of PEMDAS tells us to evaluate the parenthesis. $(4+5)=9$, simplifying the equation to $3\times9-6+7=28$. 2. There are no exponents in this problem, so we move on to multiplication. $3\times9=27$. We simplify the equation to $27-6+7=28$. 3. No division is present, so we perform addition. $(-6)+7=1$. Now your equation should be: $27+1=28$. 4. Lastly, we need to add $27$ and $1$ together ($27+1=28$) leaving us with $28=28$. By solving the left side of the equation with PEMDAS, we were able to successfully prove it is equal to the right side, giving us $28=28$.
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