Prove that the set $C = \{x \in\Bbb R : ax\le b\}$ is convex

Question

Prove that if a and b are real numbers, then the set $C = \{x \in\Bbb R : ax\le b\}$ is a convex set.

My solution so far:

1. To show that a set $C$ is convex it needs to be shown that for for every element x and y in $C$ and for every real number t between 0 and 1, $tx + (1-t)y$ is an element of $C$.
2. Let $x, y \in C$.
3. Therefore $ax\le b$ and $ay \le b$.
4. To show that the set is convex it needs to be shown that $a(tx + (1-t)y \le b$ or after some algebra that $t(ax - ay) + ay \le b.$

It seems to me that at this point I should somehow show that since both $ax \le b$ and $ay \le b$ and $t\in [0, 1]$ the inequality holds true but I'm not sure how. Am I on the right way and if I am, then what is the next step. Thanks for the help.

Answer 1

The inequalities are preserved because $t$ and $(1-t)$ are positive: $$tAx\leq tb$$ $$(1-t)Ay \leq (1-t)b$$

So: $$A(tx+(1-t)y)=tAx + (1-t)Ay\leq tb + (1-t)b = b$$

Answer 2

Hint. Rewrite $t(ax-ay) + ay$ as $(1-t)ay + tax$, now use $1-t, t \ge 0$ and $ax, ay \le b$.