Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.2 - Operations with Real Numbers and Simplifying Algebraic Expressions - Exercise Set - Page 29: 179

Answer

The simplified value is $ -7 $.

Work Step by Step

$ =\frac{9[4-(1+6)]-(3-9)^2}{5+\frac{12}{5-\frac{6}{2+1}}} $ Start from the lowest term of the denominator. $ =\frac{9[4-(1+6)]-(3-9)^2}{5+\frac{12}{5-\frac{6}{3}}} $ $ =\frac{9[4-(1+6)]-(3-9)^2}{5+\frac{12}{5-2}} $ $ =\frac{9[4-(1+6)]-(3-9)^2}{5+\frac{12}{3}} $ $ =\frac{9[4-(1+6)]-(3-9)^2}{5+4} $ $ =\frac{9[4-(1+6)]-(3-9)^2}{9} $ Now solve the numrator by using BODMAS rule. $ =\frac{9[4-7]-(3-9)^2}{9} $ $ =\frac{9[-3]-(-6)^2}{9} $ $ =\frac{-27-36}{9} $ $ =\frac{-63}{9} $ $ =-7 $
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