Solve for b given that (2b-3)/(3b+2)=(2b+1)/(3b-2)

Do you mean to solve (2b-3)/(3b+2) = (2b+1)/(3b-2)
?

Cross multiply:

(2b-3)(3b-2)
= (2b+1)(3b+2)

6b^2-4b-9b+6 =
6b^2+4b+3b+2

Collect like terms
together:

(6-6)b^2+(-4-9-4-3)b+6-2 =
0

-20b+4 = 0

20b =
4

b = 4/20

b =
1/5.

But

2b-3/3b+2=2b+1/3b-2
is different from (2n-3)/(3b+2) = (2b+1)/(3b-2)

We solve
2b-3/3b+2=2b+1/3b-2 as

Collect like terms on one side and
nubers on the other side.

2b-2b -3/3b-1/3b =
-2-2

-4/3b = -4

3b/4 =
1/4

b
=1/3.