Question 281605
I would get these both into the standard form
and solve simultaneously
{{{y = (-1/2)x + 3}}}
{{{(1/2)*x + y = 3}}}
(1) {{{x + 2y = 6}}}
and
{{{y = 2x + 8}}}
(2) {{{2x - y = -8}}}
Multiply both sides of (2) by {{{2}}}
and add (1) and (2)
(1) {{{x + 2y = 6}}}
(2) {{{4x - 2y = -16}}}
{{{5x = -10}}}
{{{x = -2}}}
and, substituting,
(1) {{{-2 + 2y = 6}}}
{{{2y = 8}}}
{{{y = 4}}}
The lines intersect at (-2,4)
here's the plot:
{{{ graph( 500, 500, -6, 8, -6, 8, (-1/2)*x + 3, 2x + 8) }}}