rewrite the equation as ...
[2x/(x-3)](x(x-3)) + [2/x](x(x-3)) = 3(x(x-3))
notice in the first term, x-3 on top cancels x-3 at the bottom, in the second term x on top cancels x at the bottom , to get ...
2x(x) + 2(x-3) = 3x^2 - 9x
2x^2 + 2x - 6 = 3x^2 - 9x
x^2 - 11x + 6 = 0
x = (11 ± √97)/2
Use the quadratic formula to find the real solutions:
2x/x-3 + 2/x = 3
I know I multiply both sides by x(x-3), but I don't know how to do that.
2 answers
thank you!
1 answer