Question 1188367


{{{3x+y=-8}}}........eq.1
{{{-2x-y=6}}}.........eq.2
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{{{-2x-y=6}}}.........eq.2, solve for {{{y}}}

{{{-2x-6=y}}}.........substitute in eq.1

{{{3x+(-2x-6)=-8}}}........eq.1, solve for {{{x}}}

{{{3x-2x-6=-8}}}
{{{x=-8+6}}}
{{{x=-2}}}

go to {{{-2x-6=y}}},substitute {{{x}}}

{{{-2(-2)-6=y}}}

{{{4-6=y}}}

{{{y=-2}}}

solution: {{{x=-2}}}, {{{y=-2}}}


{{{ drawing( 600, 600, -10, 10, -10, 10, 
circle(-2,-2,.12), locate(-2,-2,p(-2,-2)),
graph( 600, 600, -10, 10, -10, 10, -2x-6, -3x-8)) }}}