Solve the system of equations by using the elimination method:

Question

anonymous · Answer

$$4x ^{2}+3y ^{2}=12 $$

anonymous · Answer

$$5x ^{2}+6y ^{2}=30$$

blurbendy · Answer

first decide which variable you want to eliminate

anonymous · Answer

x

blurbendy · Answer

okay, in that case, the coefficients in front of x^2 need to cancel out completely.  right now 4x^2 and 5x^2 will never cancel out, but if one coefficient was -20, and the other was +20, then you can cancel them out.  how would you do this?

anonymous · Answer

multiply the top by 5 and the bottom by 4?

blurbendy · Answer

if you need a -20, you need to multiply of the equations by a negative.  either -5 or -4

blurbendy · Answer

one of the equations*

anonymous · Answer

ok say I made the top -5 and i get  $$-20x ^{2}-15y ^{2}=-60$$ what do I do next?

blurbendy · Answer

good and the bottom should be:

+ 20x^2 +24y^2 = 120

so now add the equations

-20x^2 - 15y^2 = 60
+
+ 20x^2 +24y^2 = 120

blurbendy · Answer

what do you get?

anonymous · Answer

$$9y ^{2}=60$$

blurbendy · Answer

perfect, now solve for y

anonymous · Answer

when I divide 9, it doesn't divide and if I simplify it, I get $$\frac{ 20 }{ 3 }$$

blurbendy · Answer

it might work out more nicely if we eliminate y first:

-8x^2 -6y^2 = -24

+

5x^2 +6y^2 = 30

-3x^2 = 6
x^2 = -2
x = sqrt[-2] ?

blurbendy · Answer

so x = 2i

blurbendy · Answer

now, substitute 2i back into x for one of the equations and solve for y

anonymous · Answer

so wouldn't it just be no solution?