Question 120615

Start with the given system

{{{2x-4y=-2}}}
{{{x=3y+1}}}




{{{2(3y+1)-4y=-2}}}  Plug in {{{x=3y+1}}} into the first equation. In other words, replace each {{{x}}} with {{{3y+1}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{6y+2-4y=-2}}} Distribute



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



{{{2y=-2-2}}}Subtract 2 from both sides



{{{2y=-4}}} Combine like terms on the right side



{{{y=(-4)/(2)}}} Divide both sides by 2 to isolate y




{{{y=-2}}} Divide





Now that we know that {{{y=-2}}}, we can plug this into {{{x=3y+1}}} to find {{{x}}}




{{{x=3(-2)+1}}} Substitute {{{-2}}} for each {{{y}}}



{{{x=-5}}} Simplify



So our answer is {{{x=-5}}} and {{{y=-2}}} which also looks like *[Tex \LARGE \left(-5,-2\right)]