please help me solve the problem?

Question

please work out the problem!!

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

Answer 1

Multiply both sides by (b^2-1), the LCM of the denominators
(b+3) - (b-1) = (b-2)(b+1)
4 = b^2-b-2
b^2-b-6 = 0
(b-3)(b+2) = 0
b = 3, -2

Answer 2

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

To solve this, first factor the denominator as a difference of squares.

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

To eliminate all fractions, multiply the equation by the greatest common denominator, (b - 1)(b + 1).

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

Expand and solve.

b + 3 - b + 1 = b^2 - b - 2

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

Therefore,

b = { 3, -2 }

Answer 3

(b+3) / (b^2-1) - 1 / (b+1) = (b-2)/(b-1)
(b+3) / (b^2-1) - 1 / (b+1) - (b-2) / (b-1) = 0
The lowest common multiple of the denominators is (b^2-1)
While computing,
(b+3)-(b-1)-{(b+1)(b-2)} / (b^2-1)
(b+3)-(b-1)-{b^2+b-2b-2} / (b^2-1)
(b+3 - b+1- b^2- b + 2b + 2) / (b^2 - 1)
-b^2 + b + 6 = 0
Solving this quadratic equation, by "splitting the middle term" method:
-b^2 + 3b - 2b +6 = 0
-b(b - 3) -2(b - 3) = 0
(-b - 2)(b - 3) = 0
The product of two quantities is zero. That means one of the quantitiesis zero.
Let(-b-2) = 0
-b = 2 or
b = -2
Again, let (b-3) = 0
b = 3
Hence, the value of "b" is 3 or -2....Ans
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Answer 4

((b+3)/(b^2-1)) - ((1)/(b+1)) = ((b-2)/(b-1))
Change all terms to (b+1)(b-1)
(b+3) / (b+1)(b-1) - 1(b-1)/(b+1)(b-1) = (b-2)(b+1)/(b-1)(b+1)
Multiply to eliminate the (b+1)(b-1)
b+3 - (b-1) = (b-2)(b+1)
4 = b^2 -b -2
0 = b^2 -b -6
0 = (b-3)(b+2)
b = 3, -2
Check
6/8 - 1/4 = 1/2
1/2 = 1/2 checks using 3
1/3 - 1/(-1) = -4/-3
4/3 = 4/3 checks using -2

Answer 5

((b+3)/(b^2-1)) - ((1)/(b+1)) = ((b-2)/(b-1))
((b+3)/(b -1)(b+1)) - ((1)/(b+1)) = ((b-2)/(b-1))
N.B. b^2-1 = (b-1)(b+1) difference of two squares
Multiply each of the term by (b-1)(b+1) to get
(b+3) - 1(b-1) = (b-2)(b+1)
Expanding
b+3 - b + 1 = b^2 + b - 2b - 2
4 = b^2 - b -2
Collect terms to get a quadratic equation in b:
b^2 - b - 6 = 0
Factorising to give
(b - 3)(b + 2) = 0
solving this, gives
b = -2 0r 3.

Answer 6

LHS
(b + 3) / (b - 1) (b + 1) - 1 / (b + 1)
[ (b + 3) - (b - 1) ] / [ (b - 1) (b + 1) ]
4 / [ (b - 1) (b + 1) ]

RHS
(b - 2) (b + 1) / (b - 1) (b + 1)

4 = (b - 2) (b + 1)
b² - b - 2 = 4
b² - b - 6 = 0
(b - 3) (b + 2) = 0
b = - 2 , b = 3

Answer 7

ever heard of a scientific calculator?