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Write the fraction in terms of the LCM of the denominators.

1. 5/b² , 6/ab

2. 4/(n-1)² , 3/ n(n-1)

Guest Mar 3, 2015

#2
+92808
+5

1. 5/b² , 6/ab

The LCM is ab^2

So we have

5a/ab^2 ,  6b / ab^2

2. 4/(n-1)² , 3/ [n(n-1)]

The LCM is n(n-1)^2

So we have

4n / [n(n-1)^2] , 3(n-1)/ [n(n-1)^2]

CPhill  Mar 3, 2015
#1
+5

Often, the best first step in finding the LCM is multiplying each denominator by a factor that will make them all equal (you can always reduce later).

First example:

If it's inuitive to you (and it should get there, with practice), you can see that 'b^2' is going to need an 'a' to match and 'ab' is going to need an extra 'b' to match the other side.  In order to not change the equation, you have to multiply each side by something/over-the-same-thing (because anything over itself is just one; and multiplying by one never changes anythig).  To illustrate:

(5*a) / (b^2*a) , (6*b) / (ab * b) --> 5a/ab^2 , 6b/ab^2

If you're not comfortable with that approach, you can always multiple the opposite rational by the other's entire denominator:

(5*ab) / (b^2 * ab) , (6*b^2 / (ab * b^2) --> Reduces to the same as above

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For the second, notice both have (n-1) in common, one has two of them, and the other just has an extra 'n'

4/(n-1)² , 3/ n(n-1) --> (4*n) / n(n-1)^2 , (3*(n-1) / n(n-1)^2

Guest Mar 3, 2015
#2
+92808
+5

1. 5/b² , 6/ab

The LCM is ab^2

So we have

5a/ab^2 ,  6b / ab^2

2. 4/(n-1)² , 3/ [n(n-1)]

The LCM is n(n-1)^2

So we have

4n / [n(n-1)^2] , 3(n-1)/ [n(n-1)^2]

CPhill  Mar 3, 2015