Question 147205



Start with the given system of equations:


{{{system(x-y=3,6x+8y=-38)}}}




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


{{{x-y=3}}} Start with the first equation



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



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



{{{y=(-x+3)/(-1)}}} Divide both sides by {{{-1}}}



{{{y=((-1)/(-1))x+(3)/(-1)}}} Break up the fraction



{{{y=x-3}}} Reduce




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


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




{{{6x+8highlight((x-3))=-38}}} Plug in {{{y=x-3}}} into the second equation. In other words, replace each {{{y}}} with {{{x-3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{6x+(8)(1)x+(8)(-3)=-38}}} Distribute {{{8}}} to {{{x-3}}}



{{{6x+8x-24=-38}}} Multiply



{{{14x-24=-38}}} Combine like terms on the left side



{{{14x=-38+24}}}Add 24 to both sides



{{{14x=-14}}} Combine like terms on the right side



{{{x=(-14)/(14)}}} Divide both sides by 14 to isolate x




{{{x=-1}}} Divide






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



So the first part of our answer is: {{{x=-1}}}










Since we know that {{{x=-1}}} we can plug it into the equation {{{y=x-3}}} (remember we previously solved for {{{y}}} in the first equation).




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



{{{y=(-1)-3}}} Plug in {{{x=-1}}}



{{{y=-1-3}}} Multiply



{{{y=-4}}} Combine like terms 




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



So the second part of our answer is: {{{y=-4}}}










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


So our answers are:


{{{x=-1}}} and {{{y=-4}}}


which form the point *[Tex \LARGE \left(-1,-4\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(-1,-4\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (3-1*x)/(-1), (-38-6*x)/(8) ),
  blue(circle(-1,-4,0.1)),
  blue(circle(-1,-4,0.12)),
  blue(circle(-1,-4,0.15))
)
}}} graph of {{{x-y=3}}} (red) and {{{6x+8y=-38}}} (green)  and the intersection of the lines (blue circle).