Question 130331
You can solve this by substitution, elimination, or graphing



Let's solve this system by substitution







Start with the given system of equations:


{{{system(3x-2y=12,2x+y=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


{{{3x-2y=12}}} Start with the first equation



{{{-2y=12-3x}}}  Subtract {{{3x}}} from both sides



{{{-2y=-3x+12}}} Rearrange the equation



{{{y=(-3x+12)/(-2)}}} Divide both sides by {{{-2}}}



{{{y=((-3)/(-2))x+(12)/(-2)}}} Break up the fraction



{{{y=(3/2)x-6}}} Reduce




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


Since {{{y=(3/2)x-6}}}, we can now replace each {{{y}}} in the second equation with {{{(3/2)x-6}}} to solve for {{{x}}}




{{{2x+highlight(((3/2)x-6))=8}}} Plug in {{{y=(3/2)x-6}}} into the first equation. In other words, replace each {{{y}}} with {{{(3/2)x-6}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{(2)(2x+(3/2)x-6)=(2)(8)}}} Multiply both sides by the LCM of 2. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{4x+3x-12=16}}} Distribute and multiply the LCM to each side




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



{{{7x=16+12}}}Add 12 to both sides



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



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




{{{x=4}}} Divide






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



So the first part of our answer is: {{{x=4}}}










Since we know that {{{x=4}}} we can plug it into the equation {{{y=(3/2)x-6}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(3/2)x-6}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(3/2)(4)-6}}} Plug in {{{x=4}}}



{{{y=12/2-6}}} Multiply



{{{y=6-6}}} Reduce



{{{y=0}}} Combine like terms 




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



So the second part of our answer is: {{{y=0}}}










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


So our answers are:


{{{x=4}}} and {{{y=0}}}


which form the point *[Tex \LARGE \left(4,0\right)] 









Now let's graph the two equations (if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver</a>)



From the graph, we can see that the two equations intersect at *[Tex \LARGE \left(4,0\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (12-3*x)/(-2), (8-2*x)/(1) ),
  blue(circle(4,0,0.1)),
  blue(circle(4,0,0.12)),
  blue(circle(4,0,0.15))
)
}}} graph of {{{3x-2y=12}}} (red) and {{{2x+y=8}}} (green)  and the intersection of the lines (blue circle).