## Precalculus (6th Edition)

$\color{blue}{2}$
The expressions are similar since they have the same denominator. Subtract the numerators together then copy the denominator to obtain: $=\dfrac{7x+8-(x+4)}{3x+2} \\=\dfrac{7x+8-x-4}{3x+2} \\=\dfrac{(7x-x)+(8-4)}{3x+2} \\=\dfrac{6x+4}{3x+2}$ Factor out $2$ in the numerator, then cancel the common factors to obtain: $\require{cancel} \\=\dfrac{2(3x+2)}{3x+2} \\=\dfrac{2\cancel{(3x+2)}}{\cancel{3x+2}} \\=\color{blue}{2}$