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Let's recall how do to do this with a rapid number example.
5/6 - 3/4
In this case we required a common denominator & remember that usually it's best to use the least common denominator, frequently denoted as lcd. In this case the least common denominator is 12. So we have to get the denominators of these two fractions to a 12. It is easy to do. In the first case we have to multiply the denominator by 2 to acquire 12 so we will multiply the numerator & denominator of the first fraction by 2. Recall that we've got to multiply the numerator and denominator both by the similar number as we aren't allowed to actually change the problem and it is equivalent to multiplying the fraction through 1 since (a/a)=1. . For the second term we'll need to multiply the numerator & denominator by a 3.
(5/6)-(3/4)=5(2)/6(2)-3(3)/4(3)=(10/2)-(9/12)=(10-9)/12=(1/12)
Now, the procedure for rational expressions is identical. The main complexity is finding the least common denominator. However, there is a really simple process for finding the least common denominator for rational expressions. Here is it.
1. Factor all the denominators.
2. Write each factor which appears at least once in any of the denominators. Do not write down the power which is on each factor, just write down the factor
3. Now, for each of the factor written down in the earlier step write the largest power that takes place in all the denominators containing that factor.
4. The product all the factors from the earlier step is the least common denominator.
how do you find the co=efficent when there are two brackets involved?
fgdg ggghfr hhrhfrf hfrrg jhj hjgg dear friend ghr tu vgu jyyiu ui u huik bgyuiiyts husk
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