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

$t=1$ is the only solution.
$\frac{3}{t}-\frac{t^2-2t}{t^3}=\frac{4}{t^2}$ $\frac{3t^2-t^2+2t}{t^3}=\frac{4}{t^2}$ $\frac{2t^2+2t}{t^3}=\frac{4}{t^2}$ $2t^4+2t^3=4t^3$ $2t^4-2t^3=0$ $2t^3(t-1)$ $t=0$ or $t=1$ Check: $\frac{3}{0}-\frac{0^2-2.0}{0^3}=\frac{4}{0^2}$ $0=0$ $\frac{3}{1}-\frac{1^2-2.1}{1^3}=\frac{4}{1^2}$ $4=4$ The solutions are $t=0$ or $t=1$. However, $t=0$ makes the bottom of the original equation 0, so it is not a valid answer. Thus, $t=1$ is the only solution.