Question 198900
Start with the given system of equations:

{{{system(4x-3y=11,-5x+2y=-12)}}}



{{{5(4x-3y)=5(11)}}} Multiply the both sides of the first equation by 5.



{{{20x-15y=55}}} Distribute and multiply.



{{{4(-5x+2y)=4(-12)}}} Multiply the both sides of the second equation by 4.



{{{-20x+8y=-48}}} Distribute and multiply.



So we have the new system of equations:

{{{system(20x-15y=55,-20x+8y=-48)}}}



Take note how the x coefficients are equal but opposite (so they add to zero and cancel out). 



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



{{{(20x-15y)+(-20x+8y)=(55)+(-48)}}}



{{{(20x+-20x)+(-15y+8y)=55+-48}}} Group like terms.



{{{0x+-7y=7}}} Combine like terms.



{{{-7y=7}}} Simplify.



{{{y=(7)/(-7)}}} Divide both sides by {{{-7}}} to isolate {{{y}}}.



{{{y=-1}}} Reduce.



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{{{20x-15y=55}}} Now go back to the first equation.



{{{20x-15(-1)=55}}} Plug in {{{y=-1}}}.



{{{20x+15=55}}} Multiply.



{{{20x=55-15}}} Subtract {{{15}}} from both sides.



{{{20x=40}}} Combine like terms on the right side.



{{{x=(40)/(20)}}} Divide both sides by {{{20}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



So the solutions are {{{x=2}}} and {{{y=-1}}}.



Which form the ordered pair *[Tex \LARGE \left(2,-1\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(2,-1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-11,9,
grid(1),
graph(500,500,-8,12,-11,9,(11-4x)/(-3),(-12+5x)/(2)),
circle(2,-1,0.05),
circle(2,-1,0.08),
circle(2,-1,0.10)
)}}} Graph of {{{4x-3y=11}}} (red) and {{{-5x+2y=-12}}} (green)