How do you solve (2q)/(2q+3)-(2q)/(2q-3)=12q2q+32q2q3=1?

1 Answer
Nov 27, 2016

We should find a common denominator of (2q+3)(2q-3)(2q+3)(2q3).

(2q(2q-3))/((2q+3)(2q-3))-(2q(2q+3))/((2q+3)(2q-3))=((2q+3)(2q-3))/((2q+3)(2q-3))2q(2q3)(2q+3)(2q3)2q(2q+3)(2q+3)(2q3)=(2q+3)(2q3)(2q+3)(2q3)

Combining:

(2q(2q-3)-2q(2q+3))/(4q^2-9)=(4q^2-9)/(4q^2-9)2q(2q3)2q(2q+3)4q29=4q294q29

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

(4q^2-6q)-(4q^2+6q)=4q^2-9(4q26q)(4q2+6q)=4q29

Pay attention to positives and negatives here:

-12q=4q^2-912q=4q29

4q^2+12q-9=04q2+12q9=0

Using the Quadratic Formula:

q=(-b+-sqrt(b^2-4ac))/(2a)=(-12+-sqrt(144+144))/8q=b±b24ac2a=12±144+1448

q=(-12+-12sqrt2)/8=(-3+-3sqrt2)/4q=12±1228=3±324