## Algebra: A Combined Approach (4th Edition)

(a) $\frac{1}{3}$ (b) 126 (c) 2
Substituting $x=-1$ and $y=-5$ in each of the three expressions, Part (a) $\frac{3y}{45x}=\frac{3(-5)}{45(-1)}=\frac{-15}{-45}=\frac{1}{3}$ Part (b) $x^{2}-y^{3}=(-1)^{2}-(-5)^{3}=(1)-(-125)=1+125=126$ Part (c) $\frac{x+y}{3x}=\frac{-1-5}{3(-1)}=\frac{-6}{-3}=2$