## Intermediate Algebra (12th Edition)

$(3x+7)^o=40^o ,\\ (9x-4)^o=95^o ,\\ \left( 4x+1 \right)^o=45^o$
$\bf{\text{Solution Outline:}}$ Equate the sum of the angles to $180$ and solve for $x.$ Then substitute the value of $x$ in the following given angles: \begin{array}{l}\require{cancel} (3x+7)^o ,\\ (9x-4)^o ,\\ \left( 4x+1 \right)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of the angles of a triangle is always $180^o,$ then equate the sum of the given angles to $180.$ That is, \begin{array}{l}\require{cancel} (3x+7)+(9x-4)+\left( 4x+1 \right)=180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3x+9x+4x=180-7+4-1 \\\\ 16x=176 \\\\ x=\dfrac{176}{16} \\\\ x=11 .\end{array} Substituting $x= 11$ in the angle $(3x+7)^o$ results to \begin{array}{l}\require{cancel} (3\cdot11+7)^o \\\\= (33+7)^o \\\\= 40^o .\end{array} Substituting $x= 11$ in the angle $(9x-4)^o$ results to \begin{array}{l}\require{cancel} (9\cdot11-4)^o \\\\= (99-4)^o \\\\= 95^o .\end{array} Substituting $x= 11$ in the angle $(4x+1)^o$ results to \begin{array}{l}\require{cancel} (4\cdot11+1)^o \\\\= (44+1)^o \\\\= 45^o .\end{array} Hence, the measures of the angles of the triangle are \begin{array}{l}\require{cancel} (3x+7)^o=40^o ,\\ (9x-4)^o=95^o ,\\ \left( 4x+1 \right)^o=45^o .\end{array}