It is known that there is such a number $s$ with real numbers $a,b,c,d$ that are not equal to $0$ nor $1$ and satisfy the following equations. $$ a+b+c+d=s$$
$$1/a+1/b+1/c+1/d=s$$ $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=s$$ Solve for $s$
The first part of the solution states that if $a,b,c,d$ satisfy the required conditions, so do $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$. Therefore,
$$\frac{1}{1-\frac{1}{a}}+\frac{1}{1-\frac{1}{b}}+\frac{1}{1-\frac{1}{c}}+\frac{1}{1-\frac{1}{d}}=s$$
Can someone explain to me why? (I'm still puzzled how can you do that.)
If you put $\frac 1a,\frac1b,\frac1c,\frac1d$ into first two equations you get the same two equations you started with. So if $\frac 1a,\frac1b,\frac1c,\frac1d$ aren't solutions then $a,b,c,d$ aren't solutions as well since either $$\frac 1a+\frac1b+\frac1c+\frac1d\neq s$$ Or $$\frac 1{\frac 1a}+\frac1{\frac1b}+\frac1{\frac1c}+\frac1{\frac1d}=a+b+c+d\neq s$$