## Intermediate Algebra (12th Edition)

$(5x-1)^o=64^o ,\\ (2x)^o=26^o$
$\bf{\text{Solution Outline:}}$ Equate the given complementary angles to $90$, and use the properties of equality to isolate $x.$ Then substitute the value of $x$ in the following given complementary angles: \begin{array}{l}\require{cancel} (5x-1)^o ,\\ (2x)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of complementary angles is $90^o,$ then \begin{array}{l}\require{cancel} (5x-1)+(2x)=90 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 5x+2x=90+1 \\\\ 7x=91 \\\\ x=\dfrac{91}{7} \\\\ x=13 .\end{array} Substituting $x= 13$ in the angle $(5x-1)^o$ results to \begin{array}{l}\require{cancel} (5\cdot13-1)^o \\\\= (65-1)^o \\\\= 64^o .\end{array} Substituting $x= 13$ in the angle $(2x)^o$ results to \begin{array}{l}\require{cancel} (2\cdot13)^o \\\\= 26^o .\end{array} Hence, the measures of the complementary angles are \begin{array}{l}\require{cancel} (5x-1)^o=64^o ,\\ (2x)^o=26^o .\end{array}