Question 120342



Start with the given system

{{{x-2y=-6}}}
{{{y=-x+6}}}




{{{x-2(-x+6)=-6}}}  Plug in {{{y=-x+6}}} into the first equation. In other words, replace each {{{y}}} with {{{-x+6}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{x+2x-12=-6}}} Distribute



{{{3x-12=-6}}} Combine like terms on the left side



{{{3x=-6+12}}}Add 12 to both sides



{{{3x=6}}} Combine like terms on the right side



{{{x=(6)/(3)}}} Divide both sides by 3 to isolate x




{{{x=2}}} Divide





Now that we know that {{{x=2}}}, we can plug this into {{{y=-1x+6}}} to find {{{y}}}




{{{y=-1(2)+6}}} Substitute {{{2}}} for each {{{x}}}



{{{y=4}}} Simplify



So our answer is {{{x=2}}} and {{{y=4}}} which also looks like *[Tex \LARGE \left(2,4\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(2,4\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, (-6-x)/-2, -x+6) }}} Graph of {{{x-2y=-6}}} (red) and {{{y=-x+6}}} (green)