Question 481988



Start with the given system of equations:


{{{system(6x-2y=12,2x+2y=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 second equation


{{{2x+2y=8}}} Start with the second equation



{{{2y=8-2x}}}  Subtract {{{2x}}} from both sides



{{{2y=-2x+8}}} Rearrange the equation



{{{y=(-2x+8)/(2)}}} Divide both sides by {{{2}}}



{{{y=((-2)/(2))x+(8)/(2)}}} Break up the fraction



{{{y=-x+4}}} Reduce




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Since {{{y=-x+4}}}, we can now replace each {{{y}}} in the first equation with {{{-x+4}}} to solve for {{{x}}}




{{{6x-2highlight((-x+4))=12}}} Plug in {{{y=-x+4}}} into the first equation. In other words, replace each {{{y}}} with {{{-x+4}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{6x+(-2)(-1)x+(-2)(4)=12}}} Distribute {{{-2}}} to {{{-x+4}}}



{{{6x+2x-8=12}}} Multiply



{{{8x-8=12}}} Combine like terms on the left side



{{{8x=12+8}}}Add 8 to both sides



{{{8x=20}}} Combine like terms on the right side



{{{x=(20)/(8)}}} Divide both sides by 8 to isolate x




{{{x=5/2}}} Reduce






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



So the first part of our answer is: {{{x=5/2}}}










Since we know that {{{x=5/2}}} we can plug it into the equation {{{y=-x+4}}} (remember we previously solved for {{{y}}} in the second equation).




{{{y=-x+4}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-5/2+4}}} Plug in {{{x=5/2}}}



{{{y=3/2}}} Combine like terms  (note: if you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>)




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



So the second part of our answer is: {{{y=3/2}}}










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


So our answers are:


{{{x=5/2}}} and {{{y=3/2}}}


which form the point *[Tex \LARGE \left(\frac{5}{2},\frac{3}{2}\right)]