# How do you solve (2q)/(2q+3)-(2q)/(2q-3)=1?

Nov 27, 2016

We should find a common denominator of $\left(2 q + 3\right) \left(2 q - 3\right)$.

$\frac{2 q \left(2 q - 3\right)}{\left(2 q + 3\right) \left(2 q - 3\right)} - \frac{2 q \left(2 q + 3\right)}{\left(2 q + 3\right) \left(2 q - 3\right)} = \frac{\left(2 q + 3\right) \left(2 q - 3\right)}{\left(2 q + 3\right) \left(2 q - 3\right)}$

Combining:

$\frac{2 q \left(2 q - 3\right) - 2 q \left(2 q + 3\right)}{4 {q}^{2} - 9} = \frac{4 {q}^{2} - 9}{4 {q}^{2} - 9}$

Multiplying through and disregarding the denominator since they're all equal:

$\left(4 {q}^{2} - 6 q\right) - \left(4 {q}^{2} + 6 q\right) = 4 {q}^{2} - 9$

Pay attention to positives and negatives here:

$- 12 q = 4 {q}^{2} - 9$

$4 {q}^{2} + 12 q - 9 = 0$

$q = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 12 \pm \sqrt{144 + 144}}{8}$
$q = \frac{- 12 \pm 12 \sqrt{2}}{8} = \frac{- 3 \pm 3 \sqrt{2}}{4}$