How do you solve #5/(x-3)=(2x)/(x^2-9)#?

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Answer 1 · 2018-06-11T14:32:26

#x = -5#

Note that #x# must be different from #\pm 3#. In fact, the first denominator vanished is #x=3#, while the second vanishes if #x=\pm 3#.

Note that #x^2-9=(x+3)(x-3)#, so if we multiply and divide the left hand side by #x+3# (and this is legit since #x \ne -3#) we get

#\frac{5}{x-3}*\frac{x+3}{x+3} = \frac{2x}{x^2-9}#

The left hand side can be evaluated as we noted above, so we have

#\frac{5(x+3)}{x^2-9} = \frac{2x}{x^2-9}#

We can multiply both sides by the denominator, since it's the same for both and it can't be zero:

#5(x+3) = 2x \iff 5x+15 = 2x \iff 3x = -15 \iff x = -5#