SOLUTION: 5x+1/2x+5 - x-2/3x-4 - 2 = x^2-2/6x^2+7x-20 I am not sure how to single out the x in this situation. I dont know if i should foil the right side and combine it with the

Question 891503: 5x+1/2x+5 - x-2/3x-4 - 2 = x^2-2/6x^2+7x-20

I am not sure how to single out the x in this situation. I dont know if i should foil the right side and combine it with the left side or if i should use radicals. Help me find x please? Thank you for your time in advance
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
we are assuming this is the problem, (brackets help a lot)
- - 2 =
factor the right side denominator
- - 2 =
multiply equation by (2x+5)(3x-4), cancel the denominators and you have
:
(3x-4)(5x+1) - (2x+5)(x-2) - 2(2x+5)(3x-4) = x^2 - 2
:
(15x^2+3x-20x-4) - (2x^2-4x+5x-10) - 2(6x^2+7x-20) = x^2 - 2
:
(15x^2-17x-4) - (2x^2+x-10) - (12x^2+14x-40) = x^2 - 2
removing brackets changes the signs
15x^2-17x-4 - 2x^2-x+10 - 12x^2-14x+40 = x^2 - 2
combine like terms
15x^2 - 2x^2 - 12x^2 - x^2 -17x - x - 14x - 4 + 10 + 40 + 2 = 0
yields (x^2 cancels out
-32x + 48 = 0
-32x = -48
x = 48/32
x = +1.5
:
:
You can check this for yourself, replace x with 1.5 in the original equation

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