## Algebra 1: Common Core (15th Edition)

$p=3$
Given : $\frac{5p+2}{p} = \frac{17}{p}$ This becomes : $5p+2=17$ (After multiplying both sides of the equation with the least common denominator $p$) Thus, $5p=17-2=15$ $p=15\div5=3$ Putting $p=3$ in the initial equation, Left Hand Side, $LHS=\frac{5(3)+2}{2}=\frac{17}{3}$ Right Hand Side, $RHS=\frac{17}{3}$ Hence, $LHS=RHS$