Need some guidance - Fractions

[MathJakob]

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.

 

[tiny-tim]

Hi MathJakob!

I can just cross multiply right from the beginning giving me $4(3b-2)-7(3b+2)=\frac{1}{(3b-2)(3b+2)}$

Yup! …

except that RHS should just be 1, shouldn't it?

 

[MathJakob]

Hi MathJakob!


Yup! …

except that RHS should just be 1, shouldn't it?

Why should it just be 1?

 

[tiny-tim]

because you're multiplying both sides by (3b+2)(3b-2),

and the RHS is only 1/(3b+2)(3b-2) to start with! ​

EDIT: ooh, just noticed:

your 4(3b−2)−7(3b+2) should actually have been 4(3b+2)−7(3b-2)

 

[MathJakob]

because you're multiplying both sides by (3b+2)(3b-2),

and the RHS is only 1/(3b+2)(3b-2) to start with! ​

EDIT: ooh, just noticed:

your 4(3b−2)−7(3b+2) should actually have been 4(3b+2)−7(3b-2)

oh yeh silly me D: this always happens to me when I'm doing fractions. After a while I forget even the most basic of rules... sometimes I even forget that $\frac{x}{1}$ is really just $x$ fractions are actually one of the more difficult aspects of mathematics for me...

Sometimes I just go completely brain dead, really annoying :P

 

[Curious3141]

**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.

I didn't really read the preceding, but this is definitely not right. You've completely omitted the common denominator on the LHS.

When you combine the expressions on the LHS (cross-multiply), the denominator becomes $(3b+2)(3b-2) = 9b^2 - 4$.

There's also a sign mixup in what you wrote. So what you should end up with is:

$\displaystyle \frac{4(3b+2) - 7(3b-2)}{9b^2-4} = \frac{1}{9b^2 - 4}$

Solving that is easy, just multiply both sides by $(9b^2 - 4)$. An important caveat here is that you cannot multiply both sides of an equation by zero and hope to get valid solutions. So $9b^2 - 4 \neq 0 \implies b \neq \pm\frac{2}{3}$, so you should exclude those solutions if you happen to encounter them when you work through the algebra. In this case, you won't see them.