Algebra 1: Common Core (15th Edition)

$(11,\frac{-2}{5})$
Given: $2x + 5y = 20$ $3x - 10y = 37$ Multiply both sides of the first equation with 2: $4x + 10y = 40$ $3x - 10y = 37$ Let's add the two equations to eliminate the y-terms. We get: $7x=77$ $x=11$ Now let's substitute this value of x into the first equation in order to find the value of y. $2(11) + 5y = 20$ $5y=-2$ $y=\frac{-2}{5}$ So, our solution is $(11,\frac{-2}{5})$.