1/b - 1/a = a-b/ab; This was given as a true equation, I would like to know how.?

Question

How do you go from 1/b - 1/a to a-b/ab?

Answer 1

1/b-1/a multiply nr and dr by ab,the LCD
[ab(1/b)-ab(1/a)]/ab
(a-b)/ab

Answer 2

The common denominator is ab, So multiply 1/b by a/a=1 to get a/ab and 1/a by b/b=1 to get b/ab. Then the difference is (a-b)/ab.

Look at an example to gain intuition: 1/4-1/5= (5-4)/20.

Answer 3

1/b - 1/a =

1/b (a/a) - 1/a (b/b) = (because I want a common denominator)

a/ab - b/ab which can be written more simply as a-b/ab (because they both have the same denominator.

Like someone else said, if the problem was:

1/2 - 1/3 =

1/2 (3/3) - 1/3 (2/2) = (because I want the same denominator)

3/(2*3) - 2/(2*3) = 3 - 2/ (2*3) {which is the same as a-b/ab}

This is, basically, the same format as the variable equation above, but since there are numeric values substituted in, you can solve this problem...

3 - 2/ (2*3) = 1/6

Hope that comes across clearly.....

Answer 4

Well, if it was given as a true equation then it should hold true for all values of a and b, with the possible exceptions of a = 0 or b = 0.
Consider the counter-example, where a = 1 and b = 2. By substitution:
1/2 - 1/1 = 1 - 2/(1*2)
-1/2 = 0

Whoever gave you this problem was probably yankin your chain.

Answer 5

In order to do the subtraction of fractions, they should have a common denominator. If you multiply both numarator and denominator of fraction 1/b by a and both numarator and denominator of fraction 1/a iby b, you will obtain:
a/ab - b/ab
(which is exactly equal tp 1/b-1/a, because you can cancel a from frac.1 and b from frac.2 any time)

Now you have two fractions with common denominator ab. So you can write one denominator under the line and put the two numinators above the line and obtain:
a-b/ab

Answer 6

1/b - 1/a = (a-b)/ab

let x and y represent two constants

(a-b)/ab= x/b +y/a ..............(1)

multiply through by ab

(a-b)= ax+by

compare the coefficients of a and b >>>>>>>x= 1 and y= -1

substitute values for x and y back into equation (1)

(a-b)/ab = 1/b-1/a

therefore, 1/b-1/a = (a-b)/ab is a true equation as required

Answer 7

multibly the denominator & numerator by the same term & the value doesnt change. To find a common denominator:
1/b=a/ab
&
1/a=b/ab.
Now we have
a/ab-b/ab=(a-b)/ab

Your problem statement ommitted the ( )

Answer 8

You have to get the same denominator for both (1/b) and (1/a) so you multiply by (a/a) for (1/b) and (b/b) for (1/a). Then you can set it up as what you showed

Answer 9

Common denominators: 1/b=a/ab and 1/a=b/ba; with common denominators you can now subtract to get (a-b)/ba.

Answer 10

ok it's looks confusing but it's simple. Ok if you subtract one from one it gives you zero. Which leaves you b-a or a-b. What they mean by a-b/ab there are the same answers you can write them in any form. See simple.