SOLUTION: How do I find the system of equations 3y+2z=12 and y-z=9?

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Question 175700: How do I find the system of equations 3y+2z=12 and y-z=9?
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%283y%2B2z=12%2Cy-z=9%29


2%28y-z%29=2%289%29 Multiply the both sides of the second equation by 2.


2y-2z=18 Distribute and multiply.


So we have the new system of equations:
system%283y%2B2z=12%2C2y-2z=18%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:


%283y%2B2z%29%2B%282y-2z%29=%2812%29%2B%2818%29


%283y%2B2y%29%2B%282z%2B-2z%29=12%2B18 Group like terms.


5y%2B0z=30 Combine like terms. Notice how the z terms cancel out.


5y=30 Simplify.


y=%2830%29%2F%285%29 Divide both sides by 5 to isolate y.


y=6 Reduce.


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3y%2B2z=12 Now go back to the first equation.


3%286%29%2B2z=12 Plug in y=6.


18%2B2z=12 Multiply.


2z=12-18 Subtract 18 from both sides.


2z=-6 Combine like terms on the right side.


z=%28-6%29%2F%282%29 Divide both sides by 2 to isolate z.


z=-3 Reduce.


So our answer is y=6 and z=-3.