2009 CEMC Pascal Problems/Problem 8

Problem

If $y = 3$ , the value of $\frac{y^3 + y}{y^2 - y}$ is

$\text{ (A) }\ 2 \qquad\text{ (B) }\ 3 \qquad\text{ (C) }\ 4 \qquad\text{ (D) }\ 5 \qquad\text{ (E) }\ 6$

Plugging in $y = 3$ , we get:

$\frac{3^3 + 3}{3^2 - 3} = \frac{27 + 3}{9 - 3} = \frac{30}{6} = \boxed {\textbf {(D) } 5}$

~anabel.disher

We can do the same thing as solution 1, but simplify the expression first:

$\frac{y^3 + y}{y^2 - y} = \frac{y(y^2 + 1)}{y(y - 1)} = \frac{y^2 + 1}{y - 1}$

Plugging in $y = 3$ , we get:

$\frac{3^2 + 1}{3 - 1} = \frac{9 + 1}{2} = \frac{10}{2} = \boxed {\textbf {(D) } 5}$

~anabel.disher