Question 146584

Start with the given system of equations:

{{{system(3r-2s=-4,2r+3s=32)}}}



{{{3(3r-2s)=3(-4)}}} Multiply the both sides of the first equation by 3.



{{{9r-6s=-12}}} Distribute and multiply.



{{{2(2r+3s)=2(32)}}} Multiply the both sides of the second equation by 2.



{{{4r+6s=64}}} Distribute and multiply.



So we have the new system of equations:

{{{system(9r-6s=-12,4r+6s=64)}}}



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



{{{(9r-6s)+(4r+6s)=(-12)+(64)}}}



{{{(9r+4r)+(-6s+6s)=-12+64}}} Group like terms.



{{{13r+0s=52}}} Combine like terms. Notice how the s terms cancel out.



{{{13r=52}}} Simplify.



{{{r=(52)/(13)}}} Divide both sides by {{{13}}} to isolate {{{r}}}.



{{{r=4}}} Reduce.



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{{{9r-6s=-12}}} Now go back to the first equation.



{{{9(4)-6s=-12}}} Plug in {{{r=4}}}.



{{{36-6s=-12}}} Multiply.



{{{-6s=-12-36}}} Subtract {{{36}}} from both sides.



{{{-6s=-48}}} Combine like terms on the right side.



{{{s=(-48)/(-6)}}} Divide both sides by {{{-6}}} to isolate {{{s}}}.



{{{s=8}}} Reduce.



So our answer is {{{r=4}}} and {{{s=8}}}.



Which form the ordered pair *[Tex \LARGE \left(4,8\right)].



This means that the two equations are consistent and independent.