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I am told that $x = \dfrac{2y}{1 + y}$ where $x > 2$. I proceed to use algebra to solve for $y$:

$x = \dfrac{2y}{1 + y} > 2$ where $y \not = -1$.

$\implies x(1 + y) = 2y > 2(1 + y)$ where $y \not = -1$ and $(1 + y) > 0$ $\implies y > -1$.

Since we don't know the value of $1 + y$, we have to split it into positive and negative cases. By using the case of $y > -1$, I allowed the inequality to remain the same.

But if I substitute values in for $y$, then then the equality does not hold. I am confused as to why this is the case, because it seems like I manipulated the inequality correctly?

I would greatly appreciate it if people please took the time to explain what I did incorrectly and what the correct way is.

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    You have just shown $y >-1$ is not possible. More precisely for any such $y$, you get $ 0>2$ which is absurd. Why not try the case $y<-1$ now? – 2017-02-18
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    @Macavity thanks for the response. I thought I did something wrong because the question says that there is a $y < 0$, which includes $y < -1$; therefore, I assumed $y < -1$ must be valid. – 2017-02-18
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    @Macavity It worked. Thank you for the assistance. – 2017-02-18

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