Question 228088
Start with the given system of equations:

{{{system(x-y=9,2x+3y=33)}}}



{{{3(x-y)=3(9)}}} Multiply the both sides of the first equation by 3.



{{{3x-3y=27}}} Distribute and multiply.



So we have the new system of equations:

{{{system(3x-3y=27,2x+3y=33)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(3x-3y)+(2x+3y)=(27)+(33)}}}



{{{(3x+2x)+(-3y+3y)=27+33}}} Group like terms.



{{{5x+0y=60}}} Combine like terms.



{{{5x=60}}} Simplify.



{{{x=(60)/(5)}}} Divide both sides by {{{5}}} to isolate {{{x}}}.



{{{x=12}}} Reduce.



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



{{{3x-3y=27}}} Now go back to the first equation.



{{{3(12)-3y=27}}} Plug in {{{x=12}}}.



{{{36-3y=27}}} Multiply.



{{{-3y=27-36}}} Subtract {{{36}}} from both sides.



{{{-3y=-9}}} Combine like terms on the right side.



{{{y=(-9)/(-3)}}} Divide both sides by {{{-3}}} to isolate {{{y}}}.



{{{y=3}}} Reduce.



So the solutions are {{{x=12}}} and {{{y=3}}}.



Which form the ordered pair *[Tex \LARGE \left(12,3\right)].



This means that the system is consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(12,3\right)]. So this visually verifies our answer.



{{{drawing(500,500,-5,15,-7,13,
grid(1),
graph(500,500,-5,15,-7,13,(9-x)/(-1),(33-2x)/(3)),
circle(12,3,0.05),
circle(12,3,0.08),
circle(12,3,0.10)
)}}} 


Graph of {{{x-y=9}}} (red) and {{{2x+3y=33}}} (green)