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: 38

Answer

$(3x+1)^o=40^o ,\\ (7x+49)^o=140^o$

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+1)^o ,\\ (7x+49)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of supplementary angles is $180^o,$ then \begin{array}{l}\require{cancel} (3x+1)+(7x+49)=180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3x+7x=180-1-49 \\\\ 10x=130 \\\\ x=\dfrac{130}{10} \\\\ x=13 .\end{array} Substituting $x= 13 $ in the angle $ (3x+1)^o $ results to \begin{array}{l}\require{cancel} (3\cdot13+1)^o \\\\= (39+1)^o \\\\= 40^o .\end{array} Substituting $x= 13 $ in the angle $ (7x+49)^o $ results to \begin{array}{l}\require{cancel} (7\cdot13+49)^o \\\\= (91+49)^o \\\\= 140^o .\end{array} Hence, the measures of the supplementary angles are \begin{array}{l}\require{cancel} (3x+1)^o=40^o ,\\ (7x+49)^o=140^o .\end{array}
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