Question 124410

Start with the given system

{{{3x-y=-15}}}
{{{x=y-7}}}




{{{3(y-7)-y=-15}}}  Plug in {{{x=y-7}}} into the first equation. In other words, replace each {{{x}}} with {{{y-7}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{3y-21-y=-15}}} Distribute



{{{2y-21=-15}}} Combine like terms on the left side



{{{2y=-15+21}}}Add 21 to both sides



{{{2y=6}}} Combine like terms on the right side



{{{y=(6)/(2)}}} Divide both sides by 2 to isolate y




{{{y=3}}} Divide





Now that we know that {{{y=3}}}, we can plug this into {{{x=y-7}}} to find {{{x}}}




{{{x=(3)-7}}} Substitute {{{3}}} for each {{{y}}}



{{{x=-4}}} Simplify



So our answer is {{{x=-4}}} and {{{y=3}}} which also looks like *[Tex \LARGE \left(-4,3\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(-4,3\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, (-15-3x)/(-1), (x+7)/1) }}} Graph of {{{3x-y=-15}}} (red) and {{{x=y-7}}} (green)