Question 144943
Start with the given system

{{{-x+2y=2}}}
{{{x=4-4y}}}




{{{-(4-4y)+2y=2}}}  Plug in {{{x=4-4y}}} into the first equation. In other words, replace each {{{x}}} with {{{4-4y}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{-4+4y+2y=2}}} Distribute



{{{6y-4=2}}} Combine like terms on the left side



{{{6y=2+4}}}Add 4 to both sides



{{{6y=6}}} Combine like terms on the right side



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




{{{y=1}}} Divide





Now that we know that {{{y=1}}}, we can plug this into {{{x=4-4y}}} to find {{{x}}}




{{{x=4-4(1)}}} Substitute {{{1}}} for each {{{y}}}



{{{x=0}}} Simplify



So our answer is {{{x=0}}} and {{{y=1}}} which also looks like *[Tex \LARGE \left(0,1\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(0,1\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, (2+x)/2, (x-4)/(-4)) }}} Graph of {{{-x+2y=2}}} (red) and {{{x=4-4y}}} (green)