Finding the range of equation. Any tricks?

Question

I m working on the following problem

For real numbers $a,b$, if $a+ab+b=3$, then find the range of $m=a-ab+b$. Is there any inequalities here to use?

Answer 1

Set $s=a+b$ and $t=a-b$. This is a lossless change of variables, so transform the condition: \begin{align*} 3&=a+b+ab\\ 12&=4s+(s+t)(s-t)\\ t^2&=s^2+4s-12=(s+6)(s-2) \end{align*} Therefore the condition can be satisfied only when the right side isn't negative, i.e. $s\notin(-6,2)$.

You want to find the range of $a+b-ab=a+b-(3-a-b)=2s-3$, but that's clearly $\mathbb R\setminus (-15,1)$.

Answer 2

You are given $a+ab+b=3$, which you can make $(a+1)(b+1)=4$. Let $x=a+1, y=b+1$, so $xy=4$ $m=a-ab+b=x-1-(x-1)(y-1)+y-1=-xy+2x+2y-3=2x+2y-7=2x+\frac 8x-7$
Can you find the range of that?

Answer 3

Not sure if you can get a more elementary answer, but you could see $1+a, 1+b$ are roots of $x^2-sx+4=0$, where $s^2\ge 16$. So, we have $m=(a+ab+b)-2(1+a)(1+b)+2(1+a+1+b)-2=2s-7\not \in (-15, 1)$.