# K.A.Stroud Maths revision question

deian

## Homework Statement

In 6th edition of Engineering Mathematics, Programme F2 Intro to Algebra, so far I get it all there, but I'm stumped on one question in the revision excercise, it goes like this:

((2a-3)/4b)+((3a+2)/6b)

## Homework Equations

He then goes on to break down the answers in the back:

A) 2a/4b - 3/4b + 3a/6b +2/6b
then
B) a/2b - 3/4b + a/2b + 1/3b
then
C) a/b - 5/12b

D) 1/12b*(12a-5)

## The Attempt at a Solution

I can see how the question is re-written the first time round, and then I get the second part where the first division block is divided by 2, the third by 3 and the 4th by 2. The second one can't be made smaller obviously.

But how he gets (a/b)-(5/12b) and then 1/12b*(12a-5) is above me at the moment, I know it's something really basic that i've missed. Can anyone clarify for me please?

Thanks
Deian

Homework Helper
In order to subtract the fractions a/b and 5/12b, you must have the same denominator. Since there is "b" in both denominators, that is not a problem. But there is "12" in one denominator that is not in the other. you "put" that in the other denominator, of course, by multiplying both numerator and denominator that fraction by 12: $$\frac{a}{b}= \frac{12a}{12b}$$

Now the problem is $$\frac{a}{b}- \frac{5}{12b}$$$$= \frac{12a}{12b}- \frac{5}{12b}$$$$= \frac{12a- 5}{12b}$$

That can be written as $$\frac{1}{12b}(12a- 5)$$

deian
Hi,

Thanks for the quick reply. Thanks to you I now know how they get to the final answer from the final stage.

But, how do they get from the 2nd stage:

$$\frac{a}{2b}-\frac{3}{4b}+\frac{a}{2b}+\frac{1}{3b}$$

to the 3rd, final stage:

$$\frac{a}{b}-\frac{5}{12b}$$

I like this latex thing, bit long winded but i'll get the hang of it soon, didn't think to use it first time round.

Thanks again, knowing that will complete my day. :-)