## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(6,2)$
Multiply each equation (don't forget: both sides) as denoted: $\left\{\begin{array}{ll} 5r-3s=24 & .../\times 5\\ 3r+5s=28 & .../\times 3 \end{array}\right.$ $\left\{\begin{array}{ll} 25r-15s=120 & ...\\ 9r+15s=84 & ... \end{array}\right.\quad$ adding eliminates $s$ $34r=204$ $r=\displaystyle \frac{204}{34}=6$ Back-substitute into one of the initial equations: $3r+5s=28$ $3(6)+5s=28$ $5s=28-18$ $5s=10$ $s=2$ Form an ordered pair $(r,s)$ as the solution to the system $(6,2)$