Show that the following system of linear inequalities has no solution: $x + 2y \leq 3$,$3x + 4y \geq 12$,$x \geq 0$,and $y \geq 1$.

(N/A) We are given the system of linear inequalities: $x + 2y \leq 3$,$3x + 4y \geq 12$,$x \geq 0$,and $y \geq 1$.
First,we plot the corresponding lines $x + 2y = 3$,$3x + 4y = 12$,$x = 0$,and $y = 1$ in the coordinate plane.
$1$. The line $x + 2y = 3$ passes through the points $(0, 1.5)$ and $(3, 0)$. Testing the point $(0, 0)$,we get $0 + 2(0) = 0 \leq 3$,which is true. Thus,the region satisfying $x + 2y \leq 3$ contains the origin.
$2$. The line $3x + 4y = 12$ passes through the points $(0, 3)$ and $(4, 0)$. Testing the point $(0, 0)$,we get $3(0) + 4(0) = 0 \geq 12$,which is false. Thus,the region satisfying $3x + 4y \geq 12$ does not contain the origin.
$3$. The region $x \geq 0$ represents the right half-plane (including the $y$-axis).
$4$. The region $y \geq 1$ represents the upper half-plane (including the line $y = 1$).
By plotting these regions,we observe that the shaded region for $x + 2y \leq 3$ lies below the line $x + 2y = 3$,while the region for $3x + 4y \geq 12$ lies above the line $3x + 4y = 12$. Furthermore,the constraints $x \geq 0$ and $y \geq 1$ restrict the solution to the first quadrant above $y = 1$. As shown in the figure,there is no common region that satisfies all these inequalities simultaneously. Therefore,the system has no solution (the solution set is the empty set).