Question 125976

Start with the given system

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




{{{2x+2(4x+0)=0}}}  Plug in {{{y=4x+0}}} into the first equation. In other words, replace each {{{y}}} with {{{4x+0}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{2x+8x+0=0}}} Distribute



{{{10x+0=0}}} Combine like terms on the left side



{{{10x=0-0}}}Subtract 0 from both sides



{{{10x=-0}}} Combine like terms on the right side



{{{x=(-0)/(10)}}} Divide both sides by 10 to isolate x




{{{x=0}}} Divide





Now that we know that {{{x=0}}}, we can plug this into {{{y=4x+0}}} to find {{{y}}}




{{{y=4(0)+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, 4x) }}} Graph of {{{2x+2y=0}}} (red) and {{{y=4x+0}}} (green)