- #1
MathJakob
- 161
- 5
Homework Statement
Solve for b. ##\frac{4}{3b-2}-\frac{7}{3b+2}=\frac{1}{9b^2-4}##
The Attempt at a Solution
##\frac{4}{3b-2}-\frac{7}{3b+2}=\frac{1}{(2b-2)(3b+2)}##
##(3b-2)(3b+2)\left(\frac{4}{3b-2}\right)-\left(\frac{7}{3b+2}\right)(3b-2)(3b+2)=\left(\frac{1}{(3b-2)(3b+2)}\right)(3b-2)(3b+2)##
After cancelling this messy expression I get: ##\frac{12b+8}{(3b-2)(3b+2)}-\frac{21b-14}{(3b-2)(3b+2)}=1##
Still pretty messy so I will expand and cancel further ##\frac{4}{3b-2}-\frac{7}{3b+2}=1##
Now I am stuck :/ I'm not sure if this is even correct but I don't know how to proceed.
**EDIT** I think I see what I did wrong.
I can just cross multiply right from the beginning giving me ##4(3b-2)-7(3b+2)=\frac{1}{(3b-2)(3b+2)}## but not sure what to do now... lol I'm sure this is the sort of thing I should be doing.
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