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

Solution: $(1,3)$ (checked below)
Step 1 Solve one of the equations for one of the variables. Select the first equation as the x has a coefficient of -1. $4y-x=5+2y\qquad$ ... add $x-5-2y$ $4y-2y-5=x$ $x=2y-5$ Step 2 Substitute $2y-5$ for $x$ in the other equation and solve for $y$ $3(2y-5)+7y=24\qquad$ ... (distribute, simplify) $6y-15+7y=24\qquad$ ... add $15$, simplify $13y=39\qquad$ ... divide with $13$ $y=3$ Step 3 Substitute $3$ for $y$ in $x=2y-5$. $x=2(3)-5$ $x=1$ Solution: (x,y) = $(1,3)$ Check: $\left[\begin{array}{lll} 4(3)-1\stackrel{?}{=}5+2(3) & ..... & 3(1)+7(3)\stackrel{?}{=}24 \\ 12-1=5+6 & & 3+21=24\\ 11=11 & & 24=24\\ & & \end{array}\right]$