For each answer choice, use its coordinates as $x$ and $y$ in the inequalities. Compute $x - 2y$ and $3x + y$ and check them against $4$ and $5$ with the correct inequality symbols.

In the first inequality, the symbol is $<$ (strictly less than), so equality does not count. In the second, the symbol is $\le$ (less than or equal to), so equality does count.

In Desmos, enter the two inequalities as separate expressions:

• x - 2y < 4
• 3x + y <= 5 Desmos will draw the boundary lines and shade the regions that satisfy each inequality.

Look for the region where the shadings from both inequalities overlap. That overlapping area represents all points that satisfy both inequalities at the same time.

Add each choice as a separate point in Desmos by typing (4,0), (2,2), (0,4), and (1,3) as four different expressions. Check visually which of these plotted points lies inside the overlapping shaded region from step 2; that point is the solution to the system.

A point $(x,y)$ is a solution to the system if, when you plug $x$ and $y$ into both inequalities, each inequality becomes a true statement.

So here we need a point that makes both $x - 2y < 4$ and $3x + y \le 5$ true.

For each choice, treat the coordinates as $x$ and $y$ and substitute:

• In the first inequality, compute $x - 2y$ and check if it is less than $4$.
• In the second inequality, compute $3x + y$ and check if it is less than or equal to $5$.

A point is a solution only if it passes both checks.

Test the first, second, and fourth choices:

• Choice A, $(4,0)$:

  • First inequality: $x - 2y = 4 - 2(0) = 4$. We need $x - 2y < 4$, but $4$ is not less than $4$, so this fails.
  • It also fails the second inequality: $3x + y = 3(4) + 0 = 12$, and $12 \le 5$ is false.

• Choice B, $(2,2)$:

  • First inequality: $x - 2y = 2 - 2(2) = 2 - 4 = -2$, and $-2 < 4$ is true.
  • Second inequality: $3x + y = 3(2) + 2 = 6 + 2 = 8$, and $8 \le 5$ is false, so this point is not a solution.

• Choice D, $(1,3)$:

  • First inequality: $x - 2y = 1 - 2(3) = 1 - 6 = -5$, and $-5 < 4$ is true.
  • Second inequality: $3x + y = 3(1) + 3 = 3 + 3 = 6$, and $6 \le 5$ is false, so this point is not a solution.

So A, B, and D are all eliminated because each one fails at least one inequality.

Now test choice C, $(0,4)$:

• First inequality: $x - 2y = 0 - 2(4) = -8$, and $-8 < 4$ is true.
• Second inequality: $3x + y = 3(0) + 4 = 4$, and $4 \le 5$ is true.

Since $(0,4)$ makes both inequalities true, it is the only point that is a solution to the system. The correct answer is $(0,4)$.