put it all over a common denominator, and you have
2x(2x+3) + 1(x-3) + (3x+9)
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(x-3)(2x+3)
as long as x is not 3 or -3/2, the fraction is zero when the numerator is zero. So, now you just have to solve
2x(2x+3) + 1(x-3) + (3x+9) = 0
4x^2 + 10x + 6 = 0
2(x+1)(2x+3) = 0
x = -1 is a solution
2x+3=0 is not a solution, since the fraction is then 0/0
Solve for x:
2x/(x-3) + 1/(2x+3) + 3x+9/(x-3)(2x+3) =0
1 answer
1 answer