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Solve and check!!!!!!!!!!!!!!!!!!!!!!!!!

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

asked Apr 9, 2013 in ALGEBRA 1 by andrew Scholar

1 Answer

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Solve the equation :

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

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

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

Multiply each side by ( b + 1)(b - 1)

(3b - 2)(b - 1) = (3b - 6)(b + 1)

3b² -5b +2 = 3b² - 3b -6

Cancel the common term 3b²

-5b + 2 = -3b - 6

Add 3b to each side

-2b + 2 = 0 - 6

Subtract 2 from each side

-2b + 0 = -6 - 2

-2b = -8

Divide each side by negitive 2

Therefore b = 4.

Check the equation : (3b - 2) / (b + 1) = 4 -(b + 2) / (b - 1)

Substitute b = 4 in the equation

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

10 / 5 = 4 - 6 / 3

2 = 4 - 2

2 = 2.

answered Apr 9, 2013 by diane Scholar

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