SOLUTION: Solve by the elimination method 3r-2s=-4 2r+3s=32 what is the solution of the system

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Question 146584: Solve by the elimination method
3r-2s=-4
2r+3s=32
what is the solution of the system
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Start with the given system of equations:
system%283r-2s=-4%2C2r%2B3s=32%29


3%283r-2s%29=3%28-4%29 Multiply the both sides of the first equation by 3.


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


2%282r%2B3s%29=2%2832%29 Multiply the both sides of the second equation by 2.


4r%2B6s=64 Distribute and multiply.


So we have the new system of equations:
system%289r-6s=-12%2C4r%2B6s=64%29


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


%289r-6s%29%2B%284r%2B6s%29=%28-12%29%2B%2864%29


%289r%2B4r%29%2B%28-6s%2B6s%29=-12%2B64 Group like terms.


13r%2B0s=52 Combine like terms. Notice how the s terms cancel out.


13r=52 Simplify.


r=%2852%29%2F%2813%29 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%284%29-6s=-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=%28-48%29%2F%28-6%29 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 .


This means that the two equations are consistent and independent.