Question 181876


Start with the given system of equations:


{{{system(x-5y=36,2x+y=-16)}}}




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=-16}}} Start with the second equation



{{{y=-16-2x}}}  Subtract {{{2x}}} from both sides



{{{y=-2x-16}}} Rearrange the equation



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


Since {{{y=-2x-16}}}, we can now replace each {{{y}}} in the second equation with {{{-2x-16}}} to solve for {{{x}}}




{{{x-5highlight((-2x-16))=36}}} Plug in {{{y=-2x-16}}} into the first equation. In other words, replace each {{{y}}} with {{{-2x-16}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{x+(-5)(-2)x+(-5)(-16)=36}}} Distribute {{{-5}}} to {{{-2x-16}}}



{{{x+10x+80=36}}} Multiply



{{{11x+80=36}}} Combine like terms on the left side



{{{11x=36-80}}}Subtract 80 from both sides



{{{11x=-44}}} Combine like terms on the right side



{{{x=(-44)/(11)}}} Divide both sides by 11 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=-2x-16}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-2x-16}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-2(-4)-16}}} Plug in {{{x=-4}}}



{{{y=8-16}}} Multiply



{{{y=-8}}} Combine like terms 




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



So the second part of our answer is: {{{y=-8}}}










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


So our answers are:


{{{x=-4}}} and {{{y=-8}}}


which form the point *[Tex \LARGE \left(-4,-8\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,-8\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (36-1*x)/(-5),(-16-2*x)/(1) ),
  blue(circle(-4,-8,0.1)),
  blue(circle(-4,-8,0.12)),
  blue(circle(-4,-8,0.15))
)
}}} graph of {{{x-5y=36}}} (red) and {{{2x+y=-16}}} (green) and the intersection of the lines (blue circle).