Question 182102
Start with the given system of equations:


{{{system(x+5y=-10,2x-y=2)}}}




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-y=2}}} Start with the second 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 first equation with {{{2x-2}}} to solve for {{{x}}}




{{{x+5highlight((2x-2))=-10}}} Plug in {{{y=2x-2}}} into the first 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.




{{{x+(5)(2)x+(5)(-2)=-10}}} Distribute {{{5}}} to {{{2x-2}}}



{{{x+10x-10=-10}}} Multiply



{{{11x-10=-10}}} Combine like terms on the left side



{{{11x=-10+10}}}Add 10 to both sides



{{{11x=0}}} Combine like terms on the right side



{{{x=(0)/(11)}}} Divide both sides by 11 to isolate x




{{{x=0}}} Divide






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



So the first part of our answer is: {{{x=0}}}










Since we know that {{{x=0}}} 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(0)-2}}} Plug in {{{x=0}}}



{{{y=0-2}}} Multiply



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




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



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










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


So our answers are:


{{{x=0}}} and {{{y=-2}}}


which form the point *[Tex \LARGE \left(0,-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(0,-2\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (2-2*x)/(-1), (-10-1*x)/(5) ),
  blue(circle(0,-2,0.1)),
  blue(circle(0,-2,0.12)),
  blue(circle(0,-2,0.15))
)
}}} graph of {{{2x-y=2}}} (red) and {{{x+5y=-10}}} (green)  and the intersection of the lines (blue circle).