How to prove the set $S$ is closed under the binary operation $a*b=a+b+kab$?

Question

Let $k$ be any nonzero real number and let $S$ be all real numbers except ${-1\over k}$. Let $a,b \in S$. Define * on S by $a*b=a+b+kab$. Prove that $S$ is closed under *.

All I can think to do is set $a*b$ equal to ${-1\over k}$, but I don't know where to go from there. Is that the extent of the proof?

Answer 1

Rearrange $$a+b+kab=-1/k$$ to get $$1/k+a+b+kab=0.$$ Multiply by $k$ to get $$1+ka+kb+k^2ab=0$$or $$(1+ka)(1+kb)=0$$ Hence either $a=-\frac{1}{k}$ or $b=-\frac{1}{k}$.

Answer 2

Assume $a+b+kab=-\frac{1}{k}$ divide by $ab$, and multiply by $k$
$$k^2+k\frac{1}{a}+k\frac{1}{b}+\frac{1}{ab}=0$$ or

$$(k+\frac{1}{a})(k+\frac{1}{b})=0$$