- #1

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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?

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In summary, HallsofIvy trainees find that multiplying by 90x(x-3) eliminates fractions in equations, making the solution easier to find.f

- #1

- 81

- 0

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?

- #2

- 1,755

- 1

yes, keep going

- #3

- 81

- 0

I don't have a clue what to do next

- #4

Science Advisor

Homework Helper

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2022 Award

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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

Science Advisor

Homework Helper

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NO! If x- y= 3, a

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?

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

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- #6

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- #7

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

- #8

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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 =)

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