Answer
$\color{blue}{3}$
Work Step by Step
The expressions are similar since they have the same denominator.
Subtract the numerators together then copy the denominator to obtain:
$=\dfrac{17y+3-(-10y-18)}{9y+7}
\\=\dfrac{17y+3-(-10y) -(-18)}{9y+7}
\\=\dfrac{17y+3+10y+18}{9y+7}
\\=\dfrac{(17y+10y)+(3+18)}{9y+7}
\\=\dfrac{27y+21}{9y+7}$
Factor out $3$ in the numerator, then cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{3(9y+7)}{9y+7}
\\=\dfrac{3\cancel{(9y+7)}}{\cancel{9y+7}}
\\=\color{blue}{3}$
