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Is there any method that shows how to get two equations with the same variable and different numbers?

$$\frac{a^2+a}{1+a} = \frac{1+a}{2+2a}$$
$$\frac{a}{1+a} = \frac{2+2a}{a^2+a}$$
I can also replace the numbers, e.g. it can be $(\frac{2}{3})^2+3(\frac{1}{3})^2=\frac{17}{12}$, but I don’t know how to get two equations with the same number to get rid of the number.

A:

If you take
$$\frac{a^2+a}{1+a} = \frac{1+a}{2+2a}$$
this is already a pretty easy one: Let $a=t+1$ for some real $t$; then
$$\frac{t^2+t}{2+2t} = \frac{t+t+1}{2+2(t+1)}$$
So we want to show that
$$\frac{t^2+t+1}{2+2t} = \frac{t+t+1}{2+2(t+1)}$$
and this is just
$$\frac{t^2+t+1}{2+2t} = \frac{t^2+t+1}{2+2t}$$
which is obvious.
Your second equation is harder. You want to find two $a$ that have the same ratio, so we can divide both sides by $a$; we get
$$1 + a = \frac{a}{a+1}$$
which has two solutions, corresponding to $a=-1$ and $a=-3$.

1. Field of the Invention
The present
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