Question 143588



Start with the given system of equations:


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




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}}}




{{{3x-2highlight((x-3))=3}}} Plug in {{{y=x-3}}} into the first 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.




{{{3x+(-2)(1)x+(-2)(-3)=3}}} Distribute {{{-2}}} to {{{x-3}}}



{{{3x-2x+6=3}}} Multiply



{{{x+6=3}}} Combine like terms on the left side



{{{x=3-6}}}Subtract 6 from both sides



{{{x=-3}}} Combine like terms on the right side






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



So the first part of our answer is: {{{x=-3}}}










Since we know that {{{x=-3}}} 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=(-3)-3}}} Plug in {{{x=-3}}}



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



{{{y=-6}}} Combine like terms 




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



So the second part of our answer is: {{{y=-6}}}










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


So our answers are:


{{{x=-3}}} and {{{y=-6}}}


which form the point *[Tex \LARGE \left(-3,-6\right)] 





Since the system has a solution, this means that the system is independent and consistent.







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(-3,-6\right)]. This visually verifies our answer.





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