Question 127072
Start with the given system

{{{x+y=12}}}
{{{y=2x}}}




{{{x+2x=12}}}  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.



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



{{{x=(12)/(3)}}} Divide both sides by 3 to isolate x




{{{x=4}}} Divide





Now that we know that {{{x=4}}}, we can plug this into {{{y=2x}}} to find {{{y}}}




{{{y=2(4)}}} Substitute {{{4}}} for each {{{x}}}



{{{y=8}}} Simplify



So our answer is {{{x=4}}} and {{{y=8}}} which also looks like *[Tex \LARGE \left(4,8\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(4,8\right)]. So this verifies our answer.



{{{ graph( 500, 500, -10, 10, -10, 10, 12-x, 2x) }}} Graph of {{{x+y=12}}} (red) and {{{y=2x}}} (green)