SOLUTION: Solve the system of equations using the elimination method. 2a+3b=-1 3a+5b=-2

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Question 201006: Solve the system of equations using the elimination method.
2a+3b=-1
3a+5b=-2
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%282a%2B3b=-1%2C3a%2B5b=-2%29


-3%282a%2B3b%29=-3%28-1%29 Multiply the both sides of the first equation by -3.


-6a-9b=3 Distribute and multiply.


2%283a%2B5b%29=2%28-2%29 Multiply the both sides of the second equation by 2.


6a%2B10b=-4 Distribute and multiply.


So we have the new system of equations:
system%28-6a-9b=3%2C6a%2B10b=-4%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:


%28-6a-9b%29%2B%286a%2B10b%29=%283%29%2B%28-4%29


%28-6a%2B6a%29%2B%28-9b%2B10b%29=3%2B-4 Group like terms.


0a%2Bb=-1 Combine like terms.


b=-1 Simplify.


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-6a-9b=3 Now go back to the first equation.


-6a-9%28-1%29=3 Plug in b=-1.


-6a%2B9=3 Multiply.


-6a=3-9 Subtract 9 from both sides.


-6a=-6 Combine like terms on the right side.


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


a=1 Reduce.


So the solutions are a=1 and b=-1.