Question 678302



Start with the given system of equations:


{{{system(2x-y=2,x+3y=8)}}}



Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.



So let's isolate y in the first equation


{{{2x-y=2}}} Start with the first equation



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



{{{-y=-2x+2}}} Rearrange the equation



{{{y=(-2x+2)/(-1)}}} Divide both sides by {{{-1}}}



{{{y=((-2)/(-1))x+(2)/(-1)}}} Break up the fraction



{{{y=2x-2}}} Reduce



---------------------


Since {{{y=2x-2}}}, we can now replace each {{{y}}} in the second equation with {{{2x-2}}} to solve for {{{x}}}



{{{x+3highlight((2x-2))=8}}} Plug in {{{y=2x-2}}} into the second equation. In other words, replace each {{{y}}} with {{{2x-2}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{x+(3)(2)x+(3)(-2)=8}}} Distribute {{{3}}} to {{{2x-2}}}



{{{x+6x-6=8}}} Multiply



{{{7x-6=8}}} Combine like terms on the left side



{{{7x=8+6}}}Add 6 to both sides



{{{7x=14}}} Combine like terms on the right side



{{{x=(14)/(7)}}} Divide both sides by 7 to isolate x



{{{x=2}}} Divide



-----------------First Answer------------------------------



So the first part of our answer is: {{{x=2}}}



Since we know that {{{x=2}}} we can plug it into the equation {{{y=2x-2}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=2x-2}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=2(2)-2}}} Plug in {{{x=2}}}



{{{y=4-2}}} Multiply



{{{y=2}}} Combine like terms 



-----------------Second Answer------------------------------



So the second part of our answer is: {{{y=2}}}




-----------------Summary------------------------------


So our answers are:


{{{x=2}}} and {{{y=2}}}


which form the point *[Tex \LARGE \left(2,2\right)]