# Variable of an equation is in the denominator

1. Aug 3, 2013

### mileena

1. The problem statement, all variables and given/known data

1/x - 1/a = 1/b

Solve for x

2. Relevant equations

3. The attempt at a solution

I used to be able to do this in high school, but now I am 43! Normally, I would just cross-multiply and solve for x, to get the x out of the denominator, but you cannot cross-multiply if there is more than one numerator! Does anyone know how to solve this? Thanks!

2. Aug 3, 2013

### Staff: Mentor

Hint: a*(1/a) = 1

What can you do to both sides to get rid of that 1/x?

3. Aug 3, 2013

### HallsofIvy

There are two ways to approach this:
1: Do the subtraction on the left: since the two denominators are "a" and "x", the "common denominator is ax. Multiplying numerator and denominator of the first fraction by a and the second by x, we have a/ax- x/ax= b so that (a- x)/ax= 1/b. At this point, I would multiply on both sides by abx: b(a- x)= ax so that ba- bx= ax. Can you finish this?

2. Get the x alone on one side by adding 1/a to both sides: 1/x= 1/a+ 1/b. You can add the two fractions on the right by, as before, getting a common denominator: 1/x= b/ab+ a/ab= (b+ a)/ab. Then get x by taking the reciprocal of both sides.

4. Aug 3, 2013

### mileena

Fabulous! Yes, I can do it:

ba - bx = ax

1. Factor out the b:

b(a - x) = ax

2. Distribute:

ab - bx = ax

3. Get all the x's on one side:

ab = bx + ax

4. Factor out the x:

ab = x(b + a)

5. Solve for x:

ab/(b + a) = x

Thank you! The key was finding the common denominator to make a single fraction out of the two terms (is "term" the correct word here?) or two fractions.

I just learned a new rule from the above:

If x/y = a/b, then y/x = b/a, provided that none of the terms here equal zero

Thank you for that too!

Last edited: Aug 3, 2013
5. Aug 3, 2013

### mileena

Ok, thanks for the hint. It's been a long time, I wasn't able to solve the above until I learned from HallsofIvy to find the common factor for the two denominators of the fractions on one side.

So here goes, using your method:

1/x - 1/a = 1/b

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

b/x - b/a = b/b

b/x - b/a = 1

ba/xa - bx/xa = 1

(ba - bx)/xa = 1

1(xa) = 1(ba - bx)

xa + bx = ba

x(a + b) = ab

x = ab/(a + b)

Wow! Both solutions in both of my posts match. Thank you!

6. Aug 3, 2013

### Jufro

Mileena, in your first with a solution your first two steps are unnecessary.

You said to factor and distribute (virtually undoing the factoring.

It is not wrong and you came to the right answer, I just wanted to point it out so that you know you can start the problem from step 3.

7. Aug 8, 2013

### mileena

Of course you are right Jufro! Thank you.

In my preparation the last week for the pre-calculus assessment test I will be taking in a few days, I became more comfortable with that! I don't know what I was thinking when I posted that sequence.

Also, sorry for my late reply.