# How do you solve the system 2x+3y=8 and 2x - 4y= -6?

Feb 5, 2016

$x = 1 \mathmr{and} y = 2$
$2 x + 3 y = 8$ (1)
$2 x - 4 y = - 6$ (2)
Subtracting (2) from (1) We get $7 \cdot y = 14 \mathmr{and} y = 2$ Now putting y= 2 in equation (1) We get $2 \cdot x = 8 - 3 \cdot 2 \mathmr{and} 2 x = 2 \mathmr{and} x = 1$ So $x = 1 \mathmr{and} y = 2$[ans]