The lines x+y=|a| and ax-y=1 intersect each other in the first quadrant.

Question

The lines $x+y=|a|$ and $ax-y=1$ intersect each other in the first quadrant. Then the set of all possible values of $a$ is in the interval:

Accepted Answer

Case $a>0$: $x+y=a$, $ax-y=1$. Adding: $x(1+a)=1+a \implies x=1,\ y=a-1$. For first quadrant: $y\ge0 \implies a\ge1$.

Case $a<0$: working through similarly (as in the source) leads to no valid solutions (empty set).

So $a\in[1,\infty)$.

Solutions · 1