Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.4 - Further Applications of Linear Equations - 1.4 Exercises: 34

Answer

$(9-5x)^o=49^o ,\\ (25-3x)^o=49^o$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Equate the given vertical angles, and use the properties of equality to isolate $x.$ Then substitute the value of $x$ in the following given vertical angles: \begin{array}{l}\require{cancel} (9-5x)^o ,\\ (25-3x)^o .\end{array} $\bf{\text{Solution Details:}}$ Since vertical angles have the same measure, then \begin{array}{l}\require{cancel} 9-5x=25-3x .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} -5x+3x=25-9 \\\\ -2x=16 \\\\ x=\dfrac{16}{-2} \\\\ x=-8 .\end{array} Substituting $x= -8 $ in the angle $ (9-5x)^o $ results to \begin{array}{l}\require{cancel} (9-5\cdot(-8))^o \\\\= (9+40)^o \\\\= 49^o .\end{array} Substituting $x= -8 $ in the angle $ (25-3x)^o $ results to \begin{array}{l}\require{cancel} (25-3\cdot(-8))^o \\\\= (25+24)^o \\\\= 49^o .\end{array} Hence, the measures of the vertical angles are \begin{array}{l}\require{cancel} (9-5x)^o=49^o ,\\ (25-3x)^o=49^o .\end{array}
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