Question 124727

Start with the given system

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




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



{{{-2x+9x-9=-2}}} Distribute



{{{7x-9=-2}}} Combine like terms on the left side



{{{7x=-2+9}}}Add 9 to both sides



{{{7x=7}}} Combine like terms on the right side



{{{x=(7)/(7)}}} Divide both sides by 7 to isolate x




{{{x=1}}} Divide





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




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



{{{y=0}}} Simplify



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




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



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