Question 127303
Start with the given system

{{{y=2x}}}
{{{y=-x+3}}}




{{{-x+3=2x}}}  Plug in {{{y=-x+3}}} into the first equation. In other words, replace each {{{y}}} with {{{-x+3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{-x+3=2x}}} Distribute




{{{-x=2x-3}}}Subtract 3 from both sides



{{{-x-2x=-3}}} Subtract 2x from both sides



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



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




{{{x=1}}} Divide





Now that we know that {{{x=1}}}, we can plug this into {{{y=-x+3}}} to find {{{y}}}




{{{y=-(1)+3}}} Substitute {{{1}}} for each {{{x}}}



{{{y=2}}} Simplify



So our answer is {{{x=1}}} and {{{y=2}}} which also looks like *[Tex \LARGE \left(1,2\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(1,2\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, 2x, -x+3) }}} Graph of {{{y=2x}}} (red) and {{{y=-x+3}}} (green)