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: 37

Answer

$(3x+5)^o=65^o ,\\ (5x+15)^o=115^o$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Equate the given supplementary angles to $180$, and use the properties of equality to isolate $x.$ Then substitute the value of $x$ in the following given supplementary angles: \begin{array}{l}\require{cancel} (3x+5)^o ,\\ (5x+15)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of supplementary angles is $180^o,$ then \begin{array}{l}\require{cancel} (3x+5)+(5x+15)=180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3x+5x=180-5-15 \\\\ 8x=160 \\\\ x=\dfrac{160}{8} \\\\ x=20 .\end{array} Substituting $x= 20 $ in the angle $ (3x+5)^o $ results to \begin{array}{l}\require{cancel} (3\cdot20+5)^o \\\\= (60+5)^o \\\\= 65^o .\end{array} Substituting $x= 20 $ in the angle $ (5x+15)^o $ results to \begin{array}{l}\require{cancel} (5\cdot20+15)^o \\\\= (100+15)^o \\\\= 115^o .\end{array} Hence, the measures of the complementary angles are \begin{array}{l}\require{cancel} (3x+5)^o=65^o ,\\ (5x+15)^o=115^o .\end{array}
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