SOLUTION: Solve by elimination {{{4x^2+3y^2=12}}} {{{x^2+3y^2=12}}}

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Question 140322: Solve by elimination


4x%5E2%2B3y%5E2=12
x%5E2%2B3y%5E2=12




Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given system

4x%5E2%2B3y%5E2=12
x%5E2%2B3y%5E2=12


Multiply both sides of equation 2 by -1

4x%5E2%2B3y%5E2=12
-1%28x%5E2%2B3y%5E2%29=-1%2812%29


Distribute and multiply

4x%5E2%2B3y%5E2=12
-x%5E2-3y%5E2=-12


Now add the equations together

%284x%5E2-x%5E2%29%2B%283y%5E2-3y%5E2%29=12-12


3x%5E2=0 Combine like terms


x%5E2=0 Divide both sides by 3


x=0 Take the square root of both sides



x%5E2%2B3y%5E2=12 Go back to the second equation


%280%29%5E2%2B3y%5E2=12 Plug in x=0


3y%5E2=12 Simplify


y%5E2=4 Divide both sides by 3


y=0%2B-sqrt%284%29 Take the square root of both sides


y=2 or y=-2


So when x=0, y=2 or y=-2



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Answer:

So our solutions are

(0,2) and (0,-2)