Question 182101
Start with the given system


{{{x=3y+6}}}
{{{2x+y=-16}}}





{{{2(3y+6)+y=-16}}}  Plug in {{{x=3y+6}}} into the second equation. In other words, replace each {{{x}}} with {{{3y+6}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{6y+12+y=-16}}} Distribute



{{{7y+12=-16}}} Combine like terms on the left side



{{{7y=-16-12}}}Subtract 12 from both sides



{{{7y=-28}}} Combine like terms on the right side



{{{y=(-28)/(7)}}} Divide both sides by 7 to isolate y




{{{y=-4}}} Divide





Now that we know that {{{y=-4}}}, we can plug this into {{{x=3y+6}}} to find {{{x}}}




{{{x=3(-4)+6}}} Substitute {{{-4}}} for each {{{y}}}



{{{x=-12+6}}} Multiply



{{{x=-6}}} Add



So the solutions are {{{x=-6}}} and {{{y=-4}}} which form the ordered pair *[Tex \LARGE \left(-6,-4\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(-6,-4\right)]. So this verifies our answer.



{{{ graph( 500, 500, -10, 10, -10, 10, (x-6)/3, (-16-2x)/1) }}} Graph of {{{x=3y+6}}} (red) and {{{2x+y=-16}}} (green)