## Precalculus (6th Edition)

$\{(5,0)\}$
1. Express $x$ from the first equation (solve for x). $3x-7y=15\quad/+7y$ $3x=15+7y\qquad/\div 3$ $x=\displaystyle \frac{15+7y}{3}$ 2. Substitute $x$ in the other equation and solve. $3\displaystyle \cdot\frac{15+7y}{3}+7y=15$ $15+7y+7y=15\qquad/-15$ $14y=0\qquad/\div 14$ $y=0$ 3. Back-substitute into the expression of $x$ (step 1). $x=\displaystyle \frac{15+7(0)}{3}=\frac{15}{3}=5$ 4. Write the solution as an ordered pair. Solution set: $\{(5,0)\}$