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April 18, 2015

April 18, 2015

Posted by **brit** on Monday, April 28, 2008 at 4:45pm.

please help me I'm stuck

- algebraII -
**Reiny**, Monday, April 28, 2008 at 4:57pmretype 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 -
**brit**, Monday, April 28, 2008 at 5:20pm3b-2 b+2

____=4-_____

b+1 b-1

- algebraII -
**brit**, Monday, April 28, 2008 at 6:36pm3b-2/(b+1)=4-(b+2)/b-1

- algebraII -
**Reiny**, Monday, April 28, 2008 at 7:39pmstill 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

- algebraII -

- algebraII -

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