## Intermediate Algebra for College Students (7th Edition)

$t=6$ hours
In 1 hour, pipe 1 fills $\displaystyle \frac{1}{8}$ of the (full) pool, pipe 2 fills $\displaystyle \frac{1}{24}$ of the pool. In $t$ hours, they fill $(\displaystyle \frac{t}{8}+\frac{t}{24})$ of the pool. We want t for which $\displaystyle \frac{1}{1}$ of the pool is full (the whole job). $\displaystyle \frac{t}{8}+\frac{t}{24}=1\qquad$ ... LCD=$24$ ... multiply with $24$ $3t+t=24$ $4t=24$ $t=6$ hours