Solve the equation 2/b-2 = b/ b^2-3b+2 + b/2b-2 ` `

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sciencesolve eNotes educator| Certified Educator

You need to bring the fractions to a common denominator but first, you need to write the factored form of each denominator, such that:

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

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

`2/(b - 2) = b/(b(b - 2) - (b - 2)) + b/(2(b - 1))`

`2/(b - 2) = b/((b - 2)(b - 1)) + b/(2(b - 1))`

Bringing the terms to a common denominator yields:

`2*2(b-1)/(2(b-1)(b-2)) = 2b/(2(b-1)(b-2)) + b(b - 2)/(2(b-1)(b-2))`

Equating the numerators both sides, yields:

`4(b - 1) = 2b + b(b - 2)`

`4b - 4 = 2b + b^2 - 2b => 4b - 4 = b^2 => b^2 - 4b + 4 = 0`

`(b - 2)^2 = 0 => b - 2 = 0 => b_(1,2) = 2`

Hence, evaluating the solutions to the given equation, yields `b_(1,2) = 2.`

sciencesolve eNotes educator| Certified Educator

Completing the previous answer:

Since the fraction `2/(b - 2)` and `b/(b^2 - 3b + 2)` are not valid at `b = 2` , then there are no solution to the given equation.

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