How do `5/(t^2 - t)` + `t/(t^2 - 1)` add up? 5/t2-t + t/t2-1
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calendarEducator since 2011
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You should bring the fractions to a common denominator, hence, you need to write the factored form of each denominator to check what are the common factors such that:
`5/(t^2-t) + t/(t^2-1)`
You need to factor out t to the first denominator such that:
`5/(t(t-1)) + t/(t^2-1)`
You need to convert the difference of squares, to the second denominator, into a product, using the following formula such that:
`a^2 - b^2 = (a-b)(a+b)`
Reasoning by analogy yields:
`t^2 - 1 = (t-1)(t+1)`
`5/(t(t-1)) + t/((t-1)(t+1))`
Notice that t-1 represents a common factor to both denominators, hence, the common denominator is `t(t-1)(t+1).`
You need to multiply each fraction by missing factors such that:
`5(t+1)/(t(t-1)(t+1)) + t*t/(t(t-1)(t+1)) = (5t+5+t^2)/(t(t^2-1))`
Arranging the terms to numerator yields:
`5/(t(t-1)) + t/(t^2-1) = (t^2 + 5t + 5)/(t(t^2-1))`
Hence, adding up the fractions yields `5/(t(t-1)) + t/(t^2-1) = (t^2 + 5t + 5)/(t(t^2-1)).`
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calendarEducator since 2012
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`5/(t^2 - t)` + `t/(t^2 -1)`
First you need to factorize the denominators from which you will then be able to find a common denominator:
`=5/(t(t - 1))` + `t/((t-1)(t+1))` Did you notice that we have a difference of squares?
`therefore` LCD= t(t-1)(t+1) You must multiply both by the LCD.So you need (t+1) on the left side (because you already have the t(t-1) and you need (t) on the right side as you already have the (t-1)(t+1)
`therefore =(5(t+1))/(t(t-1)(t+1))` + `((t)(t))/(t(t-1)(t+1))` Now you can add them as they have the same denominator
`therefore ` = `(5t + 5 + t^2)/ (t(t-1)(t+1))`
`therefore = (t^2 + 5t + 5)/ (t(t^2-1))`