How do you solve $\frac{2 q}{2 q + 3} - \frac{2 q}{2 q - 3} = 1$ $(2q)/(2q+3)-(2q)/(2q-3)=1$ ?

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How do you solve $\frac{2 q}{2 q + 3} - \frac{2 q}{2 q - 3} = 1$ $(2q)/(2q+3)-(2q)/(2q-3)=1$ ?

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Answer 1 · 2016-11-27T17:48:11

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

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

Combining:

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

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

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

Pay attention to positives and negatives here:

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

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

Using the Quadratic Formula:

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

$q = \frac{- 12 \pm 12 \sqrt{2}}{8} = \frac{- 3 \pm 3 \sqrt{2}}{4}$ $q=(-12+-12sqrt2)/8=(-3+-3sqrt2)/4$

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How do you solve $\frac{2 q}{2 q + 3} - \frac{2 q}{2 q - 3} = 1$ $(2q)/(2q+3)-(2q)/(2q-3)=1$ ?

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