Question 125686

Start with the given system

{{{2x+2y=0}}}
{{{y=2x}}}




{{{2x+2(2x)=0}}}  Plug in {{{y=2x}}} into the first equation. In other words, replace each {{{y}}} with {{{2x}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{2x+4x=0}}} Multiply



{{{6x=0}}} Combine like terms on the left side



{{{x=(0)/(6)}}} Divide both sides by 6 to isolate x




{{{x=0}}} Divide





Now that we know that {{{x=0}}}, we can plug this into {{{y=2x}}} to find {{{y}}}




{{{y=2(0)}}} Substitute {{{0}}} for each {{{x}}}



{{{y=0}}} Simplify



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




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



{{{ graph( 500, 500, -5, 5, -5, 5, -x, 2x) }}} Graph of {{{2x+2y=0}}} (red) and {{{y=2x}}} (green)