Question 253710


Start with the given system of equations:


{{{system(3x+y=-8,-2x-y=6)}}}




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+y=-8}}} Start with the first equation



{{{y=-8-3x}}}  Subtract {{{3x}}} from both sides



{{{y=-3x-8}}} Rearrange the equation





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


Since {{{y=-3x-8}}}, we can now replace each {{{y}}} in the second equation with {{{-3x-8}}} to solve for {{{x}}}




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




{{{-2x+3x+8=6}}} Distribute the negative



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



{{{x=6-8}}}Subtract 8 from both sides



{{{x=-2}}} Combine like terms on the right side






-----------------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=-3x-8}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-3x-8}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-3(-2)-8}}} Plug in {{{x=-2}}}



{{{y=6-8}}} 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)] 









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(-2,-2\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  grid(1),
  graph(500, 500, -10,10,-10,10, (-8-3*x)/(1), (6--2*x)/(-1) ),
  blue(circle(-2,-2,0.1)),
  blue(circle(-2,-2,0.12)),
  blue(circle(-2,-2,0.15))
)
}}} graph of {{{3x+y=-8}}} (red) and {{{-2x-y=6}}} (green)  and the intersection of the lines (blue circle).