Minimum value of a linear program

Question

$x$ and $y$ are real numbers such that $ x\ge 0$ and $y \ge 0$.

If $$ x + y \le 5$$ $$ x + 2y \ge 8$$

Then what is the minimum value of $5x+y$?

There is something wrong in my approach. I write the first inequality as $$ 5 \ge x + y$$ Adding with the second inequality yields, $$5 + x + 2y \ge x + y + 8$$ Or $$ y \ge 3$$ Thus $$ 9y \ge 27$$ Multiplying the second inequality by $5$ we obtain $$ 5x + 10y \ge 40$$ $$9y \ge 27$$ Hence $5x+y \ge 13$

Answer 1

Our conditions give an interior of $\Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$

The system $x=0$ and $x+2y=8$ gives $C(0,4)$;

The system $x=0$ and $x+y=5$ gives $B(0,5)$

and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$

Let $f(x,y)=5x+y.$

Thus, $$\min_{x\geq0,y\geq0,x+y\leq5,x+2y\geq8}f=\min\{f(0,5),f(0,4),f(2,3)\}=f(0,4)=4.$$