Show, using the definition of convex set, that the set $S=\{(x_1,x_2) | 3x_1-5x_2\geq 15\}$ is convex.
I have started the proof by letting $y=(y_1,y_2),$ $z=(z_1,z_2) \in S$ and letting $t \in (0,1)$. I know we want to show that $ty+(1-t)z \in S$. I have plugged in $y,z$ in the equation, but am not sure how this shows that the set is convex.
Suppose that $y,z\in S$. Then $3y_1-5y_2\geq 15$ and $3z_1-5z_2\geq 15$. Let $t\in(0,1)$ and consider $ty+(1-t)z$. Note that I didn't write $ty+(1-t)z\in S$ as in the question, because the $\in S$ part is what must be shown.
Observe that $ty+(1-t)z=(ty_1+(1-t)z_1,ty_2+(1-t)z_2)$. Consider $$ 3(ty_1+(1-t)z_1)-5(ty_2+(1-t)z_2)=t(3y_1-5y_2)+(1-t)(3z_1-5z_2). $$ Since $3y_1-5y_2\geq 15$ and $3z_1-5z_2\geq 15$, we can substitute these inequalities into above to get that $$ t(3y_1-5y_2)+(1-t)(3z_1-5z_2)\geq 15t+(1-t)15=15. $$ Therefore $ty+(1-t)z\in S$.