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

$(\displaystyle \frac{49}{11},-\frac{12}{11})$
If we multiply the first equation with 2, the terms containining $y$ will be opposite in the additive sense. \begin{aligned}10x+6y&=38\\ x-6y&=11\end{aligned} Add the equations, $11x=49\qquad$ ... and solve for $x$ $x=\displaystyle \frac{49}{11}$ Substitute into one of the initial equations: $5x+3y=19$ $5(\displaystyle \frac{49}{11})+3y=19$ $3y=19-\displaystyle \frac{245}{11}$ $3y=\displaystyle \frac{209-245}{11}$ $3y=\displaystyle \frac{-36}{11}$ $y=-\displaystyle \frac{12}{11}$ Form an ordered pair $(x,y)$ as the solution to the system $(\displaystyle \frac{49}{11},-\frac{12}{11})$