# Urgent Help needed

1. May 28, 2008

### Paulo2014

The difference between two positive numbers is 3.
The difference between their reciprocals is 1/90
What are the two numbers?

I worked out that:
x-y=3 and
1/x-1/y= 1/90
Is that right so far?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 29, 2008

### rocomath

yes, keep going

3. May 29, 2008

### Paulo2014

I don't have a clue what to do next

4. May 29, 2008

### malawi_glenn

why dont you just make a guess?

When you have two equations of two unkowns, the strategy is to express one of the unkowns as a "function" of the other unknown.

e.g

y+2x=9 gives: y = 9 -2x

Then substitue that into the other equation.

5. May 29, 2008

### HallsofIvy

Staff Emeritus
NO! If x- y= 3, a positive number, then x must be greater than y. But taking reciprocals reverses order. If x> y then 1/x< 1/y so 1/x- 1/y is negative. Assuming that x is the larger of the two numbers so that x- y= 3, then you must have 1/y- 1/x= 1/90.

Now you know that y= x- 3 so 1/(x- 3)- 1/x= 1/90. I would recommend multiplying that every term in that equation by 90x(x-3) in order to get rid of the fractions.

By the way, it is always a good idea to start a problem like this, NOT by saying "x- y= 3" but by defining the variables: "Let x and y be the two numbers, with $x\ge y$".

Last edited: May 29, 2008
6. May 30, 2008

### Paulo2014

why would I need to multiply everything by 90x(x-3)? I understand about the 1/y- 1/x= 1/90 part but I don't understand anything else....

7. May 30, 2008

### konthelion

You multiply by 90x(x-3) in order to cancel the terms in the denominator, making it easier to solve for x.

8. May 31, 2008

### mistermath

OK, so far you should understand that the 2 equations you are working with are:
a: x - y = 3
b: 1/y - 1/x = 1/90

What *I* would do if I'm a rookie is multiply equation b by x*y*90 since that is the LCM of the denominators and I'm scared of fractions. This would give me:
b*: 90x - 90y = xy

at this point you can use the substitution method to solve.

HallsofIvy had a slightly different approach which comes with training. First he solved eqn a for y. ie a*: y = x -3. Then he substituted this into equation b to get:
b*: 1/(x-3) - 1/x = 1/90

now the LCM of the denomiators is x * (x-3) * 90. If you mulitply by that you get:
b**: 90x - 90(x-3) = x(x-3)

If you follow the first method, you'll see that both methods turn out to be the same, except that my way is easier because I got rid of all fractions before doing any work. After some "training" though, you get used to the fractions, and just solve without getting rid of them the second you see them.

I think the biggest lesson to take away from this is be sure to label your variables at the beginning! So I'm double emphasizing it =)