Prove the set $M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$ is convex 
Let $\alpha\gt 0$, $\gamma\gt 0$, and $\beta\gt 0$ be real numbers. Let 
  $$M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$$
  Prove $M$ is a convex set. Prove that $M$ is bounded. What does this set resemble (in economics)?

Attempt:
If $(x_1,x_2),(y_1,y_2)\in M$ we get
$$\begin{align*}
\alpha x_1 + \gamma x_2&\leq \beta\\
\alpha y_1 + \gamma y_2 &\leq \beta
\end{align*}$$
We want to prove
$$\alpha(ax_1 + (1-a)y_1) + \gamma(ax_2 + (1-a)y_2)\leq \beta.$$
The question is how do I prove this inequality?
 A: Algebra! (pronounced like Jon Lovitz's Master Thespian character)
$$\begin{align*}
\alpha(ax_1 + (1-a)y_1) + \gamma(ax_2+(1-a)y_2) &= \alpha ax_1 + \gamma ax_2 + \alpha(1-a)y_1 + \gamma(1-a)y_2\\
&= a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\\
&\leq a\beta + (1-a)\beta.
\end{align*}$$
A: Same thing Arturo posted in more detail:
We know that the below two inequalities on the far left are true. So lets use them to prove the one we need to prove$[α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤β]$:
$αx_1+γx_2≤β\;\;|*a\Rightarrow a(\alpha x_1+\gamma x_2)\leq a\beta$
$αy_1+γy_2≤β\;\;|*(1-a)\Rightarrow (1-a)(αy_1+γy_2)\leq (1-a)\beta$
Now add the inequalities on the far right side and we get:
$$a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq a\beta+(1-a)\beta$$
After expanding the expressions in parenthesis on the LHS and rearranging the terms we get:
$$α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤a\beta+(1-a)\beta$$
Which almost looks exactly like the one we need to prove. The RHS after expanding: 
$$a\beta+(1-a)\beta=a\beta +\beta -a\beta =\beta \Rightarrow$$ 
$$\Rightarrow a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq\beta$$
Which is what we needed to show.
2&3 questions:
This set M={$x∈ℝ^2_+∣αx_1+γx_2≤β$} looks like a budget constraint and is bounded by:
if $x_1=0;\;$ $\gamma x_2\leq \beta$;$\;\;x_2\leq \frac{\beta}{\gamma}$
if $x_2=0;\;$ $\alpha x_1\leq \beta$;$\;\;x_1\leq \frac{\beta}{\alpha}$

