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Question: Find values for $a$, $b$, $n$ such that: $\frac{-ab(n+1)x-(an+b)y}{(a+bn)x+(n+1)y} = \frac{10x-4y}{3x-y}$

Attempt: Tried using the following relationships by comparing coefficients and solving for $a$, $b$, $n$:

$-ab(n+1)=10, an+b=4, a+bn=3, n+1=-1$

This gives no values of $a$, $b$, $n$ that satisfy all four equations.

Dividing the numerator and denominator by $-(n+1)$ instead which gives $\frac{abx+\frac{an+b}{n+1}y}{-\frac{a+bn}{n+1}x-y}$ and then comparing coefficients gives valid $a$, $b$, $n$. Why is this? I understand that this ensures the correct coefficient for $y$ in the denominator for all values, but I don't see how that is significant and how the first method does not also lead to the correct solution.

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HINT

The original equations are solved if there is a number $k$ such that $$-ab(n+1)=10k,an+b=4k, a+bn=3k, n+1=-k.$$

You can use the fourth of these equations to eliminate $k$ and then you'll get a solution.

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The first time around, you had four unknowns and four equations. The second time around, you only had three equations. So the second method was likely on general principles to be better able to give you an answer, because it involved solving a less strict set of simultaneous equations.

The first method forced $n$ to be something specific; the second method eliminated it from the set of equations, and hence allowed it to vary.

Of course, you might still not have found all the solutions! You'll need to prove that you've found all the solutions (if it's true and if that's something you want to prove). The solution set you got from the first method is a valid solution set; it's just smaller than the solution set you got from the second method. But maybe there's another transformation you could do to expand even the second method's solution set. (I don't know; I haven't tried.)

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hint

You need three equations $$\lim_{x\to +\infty}f(x,y)=\frac{-ab(n+1)}{a+bn}=\frac{10}{3}$$ $$\lim_{y\to+\infty}f(x,y)=\frac{-(an+b)}{n+1}=4$$

and, if $ y=3x$, the denominator should be zero:

$$a+bn+3(n+1)=0$$

for example, you will find that $ab=10$.

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