Solve the following equation: 2/3-x + 3/x+3= 6x/x^2-9
2/(3-x) + 3/(x+3)= 6x/(x^2-9) multiple (x^2 - 9) on both sides: -2(x + 3) + 3(x - 3) = 6x -2x - 6 + 3x - 9 = 6x -5x = 15 x = -3 while when x = -3, (x+3) = 0; (x^2-9) = 0 We know that denominator can not be zero. Therefore, there is no solution for the equation 2/(3-x) + 3/(x+3)= 6x/(x^2-9).
Question
Updated 4/12/2014 12:49:44 AM
Rating
3
2/(3-x) + 3/(x+3)= 6x/(x^2-9)
multiple (x^2 - 9) on both sides:
-2(x + 3) + 3(x - 3) = 6x
-2x - 6 + 3x - 9 = 6x
-5x = 15
x = -3
while when x = -3, (x+3) = 0; (x^2-9) = 0
We know that denominator can not be zero.
Therefore, there is no solution for the equation 2/(3-x) + 3/(x+3)= 6x/(x^2-9).
Confirmed by andrewpallarca [4/12/2014 12:52:32 PM]

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