# algebraII

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3b-2/b+1=4-b+2/b-1

• algebraII -

retype with brackets to establish the correct order of operation.

did you mean
(3b-2)/b+1 = (4-b+2)/(b-1) or
3b - 2/(b+1) = 4-b + 2/b-1 or ...

• algebraII -

3b-2 b+2
____=4-_____
b+1 b-1

• algebraII -

3b-2/(b+1)=4-(b+2)/b-1

• algebraII -

still not totally clear, I will assume you mean
(3b-2)/(b+1)=4-(b+2)/(b-1)

so the LCD is (b-1)(b-1), let's multiply each term by that

(3b-2)(b-1) = 4(b-1)(b+1) - (b+2)(b+1)
for which I go
b = 4

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