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solve the equation. `2/(b-2) = b/(b^2 - 3b + 2) + b/(2b-2)`
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`2/(b-2) = b/(b^2 - 3b + 2) + b/(2b-2)`
Factor each denominator in order to be able to find the least common denominator:
`2/(b-2) = b/((b-1)(b-2)) + b/(2(b-1))`
The common denominator is 2(b-1)(b - 2).
Multiply both sides of the equation by the common denominator:
`2(b-1)(b - 2) * 2/(b-2) = 2(b-1)(b-2) * b/((b-1)(b-2)) + 2(b-1)(b-2) * b/(2(b-1))`
`4(b-1) = 2b + b(b-2)`
`4b-4 = 2b + b^2 - 2b`
`b^2 - 4b + 4 = 0`
`(b-2)^2 = 0`
`b = 2`
However, b = 2 makes the denominator in two of the fractions in the original equation 0, so it cannot be a solution. Thus, the equation has no solutions.
Posted by ishpiro on May 31, 2013 at 4:43 PM (Answer #1)
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