Question 316755


Start with the given system of equations:


{{{system(2x+y=12,3x-y=13)}}}




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=12}}} Start with the first equation



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



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



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


Since {{{y=-2x+12}}}, we can now replace each {{{y}}} in the second equation with {{{-2x+12}}} to solve for {{{x}}}




{{{3x-highlight((-2x+12))=13}}} Plug in {{{y=-2x+12}}} into the second equation. In other words, replace each {{{y}}} with {{{-2x+12}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{3x+2x-12=13}}} Distribute the negative



{{{5x-12=13}}} Combine like terms on the left side



{{{5x=13+12}}}Add 12 to both sides



{{{5x=25}}} Combine like terms on the right side



{{{x=(25)/(5)}}} Divide both sides by 5 to isolate x




{{{x=5}}} Divide






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



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










Since we know that {{{x=5}}} we can plug it into the equation {{{y=-2x+12}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-2x+12}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-2(5)+12}}} Plug in {{{x=5}}}



{{{y=-10+12}}} Multiply



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




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



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










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


So our answers are:


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


which form the point *[Tex \LARGE \left(5,2\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(5,2\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (12-2*x)/(1), (13-3*x)/(-1) ),
  blue(circle(5,2,0.1)),
  blue(circle(5,2,0.12)),
  blue(circle(5,2,0.15))
)
}}} graph of {{{2x+y=12}}} (red) and {{{3x-y=13}}} (green)  and the intersection of the lines (blue circle).