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{ab(a-b)(a-c)} / {ac(b-a)(b-c)}, Simplify.

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lucky092569 | (Level 3) Honors

Posted January 27, 2009 at 4:52 PM via web

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{ab(a-b)(a-c)} / {ac(b-a)(b-c)}, Simplify.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted January 27, 2009 at 6:16 PM (Answer #1)

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First of all, note that you have the same letter "a" at numerator and denominator, same time, so you can simplify it. The result will be:

[b(a-b)(a-c)]/[c(b-a)(b-c)]

After that, multiply the fraction with "-1" value so that, at numerator, (a-b) will become (b-a). But, note that (b-a) paranthesys is also at denominator, so you can simplify it too. The result will be, after simplifying action:

-[b(a-c)]/c(b-c)]

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neela | High School Teacher | (Level 3) Valedictorian

Posted August 18, 2009 at 2:21 PM (Answer #2)

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To simplify {ab(a-b)(a-c)}/{ac(b-a)(b-c)}

Observe that both numerator and denominators can be expressed as below:

Numerator factors:     a*b*(a-b)(a-c)

Denominator factors:  a*c*(a-b)(-1)*(b-c).

Threfore, the highest common factor(HCF) of Numerator and Denominator =a(a-b).

Therfore, we can divide by the the HCF both numerator and denominator to get the simplified expresion of the given expression:

b(a-c)/{c(-1)(b-c)}, which is equivalent to

b(a-c)/{c(c-b)} or

b(c-a)/{c(c-b)}

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revolution | College Teacher | (Level 1) Valedictorian

Posted July 31, 2010 at 9:08 PM (Answer #3)

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{ab(a-b)(a-c)} / {ac(b-a)(b-c)}

First, divide this equation throughout by (a), as a is present in both side, so it turns out to be:

(b(a-b)(a-c)) / (c(b-a)(b-c))

Then, change (a-b) into (b-a), to be in the same form as the denominator Remember to put the negative sign in front of the numerator:

-(b(b-a)(a-c))/ (c(b-a)(b-c))

Divide b-a throughout top and bottom

-(b(a-c))/(c(b-c)) //

 

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atyourservice | Student, Grade 10 | (Level 3) Valedictorian

Posted July 30, 2014 at 10:52 PM (Answer #4)

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`{ab(a-b)(a-c)} / {ac(b-a)(b-c)}`

They both share a on the top and bottom you should first divide by a to cancel out a, that would leave you:

`(b(a-b)(a-c)) / (c(b-a)(b-c))`

Now switch the order of b-a to -a+b, 

`(b(a-b)(a-c))/ (c(-a+b)(b-c))`

Now factor out a negative to make the denominator and the nominator the same:

`(b(a-b)(a-c))/ (-c(a-b)(b-c))`

now divide by a-b to cancel them out

you should be left over with:

`(b(a-c))/(-c(b-c))`

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