Question 172445



Start with the given system of equations:


{{{system(-3x+y=-3,2x-5y=-11)}}}




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=-3}}} Start with the first equation



{{{y=-3+3x}}} Add {{{3x}}} to both sides



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




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


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




{{{2x-5highlight((3x-3))=-11}}} Plug in {{{y=3x-3}}} into the second equation. In other words, replace each {{{y}}} with {{{3x-3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{2x+(-5)(3)x+(-5)(-3)=-11}}} Distribute {{{-5}}} to {{{3x-3}}}



{{{2x-15x+15=-11}}} Multiply



{{{-13x+15=-11}}} Combine like terms on the left side



{{{-13x=-11-15}}}Subtract 15 from both sides



{{{-13x=-26}}} Combine like terms on the right side



{{{x=(-26)/(-13)}}} Divide both sides by -13 to isolate x




{{{x=2}}} Divide






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




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



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



{{{y=6-3}}} Multiply



{{{y=3}}} Combine like terms 




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



So the second part of our answer is: {{{y=3}}}










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


So our answers are:


{{{x=2}}} and {{{y=3}}}


which form the point *[Tex \LARGE \left(2,3\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(2,3\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (-3--3*x)/(1), (-11-2*x)/(-5) ),
  grid(1),
  blue(circle(2,3,0.1)),
  blue(circle(2,3,0.12)),
  blue(circle(2,3,0.15))
)
}}} graph of {{{-3x+y=-3}}} (red) and {{{2x-5y=-11}}} (green)  and the intersection of the lines (blue circle).