Question 499795



Start with the given system of equations:


{{{system(2x-y=2,3x-2y=11)}}}




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}}}




{{{3x-2highlight((2x-2))=11}}} 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.




{{{3x+(-2)(2)x+(-2)(-2)=11}}} Distribute {{{-2}}} to {{{2x-2}}}



{{{3x-4x+4=11}}} Multiply



{{{-x+4=11}}} Combine like terms on the left side



{{{-x=11-4}}}Subtract 4 from both sides



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



{{{x=(7)/(-1)}}} Divide both sides by -1 to isolate x




{{{x=-7}}} Divide






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



So the first part of our answer is: {{{x=-7}}}










Since we know that {{{x=-7}}} 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(-7)-2}}} Plug in {{{x=-7}}}



{{{y=-14-2}}} Multiply



{{{y=-16}}} Combine like terms 




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



So the second part of our answer is: {{{y=-16}}}










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


So our answers are:


{{{x=-7}}} and {{{y=-16}}}


which form the point *[Tex \LARGE \left(-7,-16\right)]