## Intermediate Algebra (12th Edition)

$(3x+1)^o=40^o ,\\ (7x+49)^o=140^o$
$\bf{\text{Solution Outline:}}$ 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}