High School Math contest division question

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SUMMARY

The math contest division question involves finding the values of x and y given specific conditions about their division. The equations derived from the problem are x/y = 3 + 7/y and y/x = Q + 12/x. By manipulating these equations, the final values determined are x = 43 and y = 12. The assumption that x, y, and Q are natural numbers simplifies the solution to a unique answer.

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timelesstrix0
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The question is : If x > 0 and I divide x by y, the quotient is 3 and remainder is 7. If I divide y by x the remainder is 12. what is the value of x?

So far I used long division quotient form to make 2 equations...

1. x/y = 3 + 7/y
2. y/x = Q(quotient) + 12/x

so 1st equation i solve for y and get y = x/3 - 7/3
2nd equation i solve for y and get y = Q(x) + 12the final answers that the teacher told me were that x = 43 and y = 12
can anyone explain how i can solve this please?
 
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I would multiply the first equation by y and the second equation by x to get rid of fractions. Afterwards you can plug the expression for x from the first equation into the second equation and look for possible values of y (in both equations).
 
If you have three variables then you should also have three equations.
Here, you have three variables(x, y, and Q) but two equations. Plz make sure the value of Q is given or not.
 
Last edited by a moderator:
Deepak suwalka said:
If you have tree variables then you should also have three equations.
Here, you have three variables(x, y, and Q) but two equations. Plz make sure the value of Q is given or not.
There are only two equations, but there is an assumption that x,y,Q ∈ ℕ, which limits the problem to a single solution.
 

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