Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 1 - Expressions, Equations, and Inequalities - 1-3 Algebraic Expressions - Practice and Problem-Solving Exercises - Page 24: 62

Answer

a) $18$ b) $18$ c) In both methods, we used the correct properties of real numbers and followed the order of operations, and since we evaluated the same or equivalent expressions, its value for $x = 3$ must be the same regardless of the method used.

Work Step by Step

a) To evaluate, we plug in $3$ to $x$: $2(2x^{2}–x) –3(x^{2}–x) + x^{2}–x$ = $2(2(3)^{2}–3) –3(3^{2}–3) + 3^{2}–3$ = $2(2(9)–3) –3(9–3) + 9–3$ = $2(18–3) –3(6) + 6$ = $2(15) – 18 + 6$ = $30 – 18 + 6$ = $18$ b) Simplifying first, we get: $2(2x^{2}–x) –3(x^{2}–x) + x^{2}–x$ = $4x^{2}–2x –3x^{2}+3x+ x^{2}–x$ = $2x^{2}$ Now we plug in $3$ to $x$: $2(3)^{2} = 2(9) = 18$
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