# Discrete Math Help

Discrete Math Help!!!

Here is the problem:

Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations:

$$\dfrac{xy}{x+y}=a$$ and $$\dfrac{xz}{x+z}=b$$ and $$\dfrac{yz}{y+z}=c$$.

Is x rational? If so, express it as a ratio of two integers.

I have calculated that $$x=\dfrac{-(bz-xz)}{b}$$. I am inclined to answer no, since x, y, or z could be irrational.

Any help would be appreciated.

hint:

first modify your equation to solve for x explicitly in terms of b and z ( x on one side , b and z terms on the other ).
then use the other two equations to come up with an expression for z and substitute that into your original equation.
Your goal is to try to solve for x in terms of a,b,c only... eliminating any z or y

Post your steps if you run into more trouble
Regards,
-MS

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Ok, I solved and got this:

$$x=\dfrac{ay}{y-a}$$ and $$x=\dfrac{bz}{z-b}$$ and $$y=\dfrac{cz}{z-c}$$.

I then solved for z and got $$z=\dfrac{aby}{(a-b)y+ab}$$. I just cancelled the y's and was left with $$z=\dfrac{ab}{(a-b)+ab}$$.

I plugged this in for z in the second equation, and got $$z=\dfrac{b \dfrac{ab}{(a-b)+ab}}{\dfrac{ab}{(a-b)+ab}-b}$$. This simplifies to $$\dfrac{ab}{b-ab}$$.

If this answer is correct, I would say that yes, x is indeed rational. Am I correct?

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Where did the x come from in the third equation (top right) , and where did the y go? That completely changes the problem so the rest is all wrong.

Also, for
$$z=\dfrac{aby}{(a-b)y+ab}$$ you can't cancel the y's, because y is not a factor of the denominator.
If the denominator was instead $(a-b)y + aby$ then you could factor out the y which could become $y((a-b) + ab)$ and the y's could cancel
(do you see the difference? )

Another thing: in your result,,, you solved for z (although incorrectly). The equation asks if X is rational... so how does knowing whether or not Z is rational help you?
edit:
based on your equation relating x to z you could use the fact that z is rational to prove that x is rational providing you knew that theorem, but this seems like more trouble than its worth why not just solve for x. /edit

Your going to have to solve for x in terms of y or z, and then solve for either y or z using the other equations to come up with an equation in terms of x, a, b, and c only

-MS

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