How do you solve #(2q)/(2q+3)-(2q)/(2q-3)=1#?
1 Answer
We should find a common denominator of
#(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#
