How do you solve #(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 #(2q+3)(2q-3)#.

#(2q(2q-3))/((2q+3)(2q-3))-(2q(2q+3))/((2q+3)(2q-3))=((2q+3)(2q-3))/((2q+3)(2q-3))#

Combining:

#(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:

#(4q^2-6q)-(4q^2+6q)=4q^2-9#

Pay attention to positives and negatives here:

#-12q=4q^2-9#

#4q^2+12q-9=0#

Using the Quadratic Formula:

#q=(-b+-sqrt(b^2-4ac))/(2a)=(-12+-sqrt(144+144))/8#

#q=(-12+-12sqrt2)/8=(-3+-3sqrt2)/4#