We are given a system of inequalities: $y \leq -2x + 6$ $x - y < 6$ We need to determine which of the given pairs of $(x, y)$ are solutions to this system. We will test each option by substituting the $x$ and $y$ values into both inequalities and checking if both inequalities are satisfied.

1. Problem Description

We are given a system of inequalities:

$y \leq -2x + 6$

$x - y < 6$

We need to determine which of the given pairs of $(x, y)$ are solutions to this system. We will test each option by substituting the $x$ and $y$ values into both inequalities and checking if both inequalities are satisfied.

2. Solution Steps

A. $x = 0, y = 0$

$0 \leq -2(0) + 6 \Rightarrow 0 \leq 6$ (True)

$0 - 0 < 6 \Rightarrow 0 < 6$ (True)

Since both inequalities are true, $(0, 0)$ is a solution.

B. $x = -5, y = -15$

$-15 \leq -2(-5) + 6 \Rightarrow -15 \leq 10 + 6 \Rightarrow -15 \leq 16$ (True)

$-5 - (-15) < 6 \Rightarrow -5 + 15 < 6 \Rightarrow 10 < 6$ (False)

Since one inequality is false, $(-5, -15)$ is not a solution.

C. $x = 4, y = -2$

$-2 \leq -2(4) + 6 \Rightarrow -2 \leq -8 + 6 \Rightarrow -2 \leq -2$ (True)

$4 - (-2) < 6 \Rightarrow 4 + 2 < 6 \Rightarrow 6 < 6$ (False)

Since one inequality is false, $(4, -2)$ is not a solution.

D. $x = 3, y = 0$

$0 \leq -2(3) + 6 \Rightarrow 0 \leq -6 + 6 \Rightarrow 0 \leq 0$ (True)

$3 - 0 < 6 \Rightarrow 3 < 6$ (True)

Since both inequalities are true, $(3, 0)$ is a solution.

E. $x = 10, y = 0$

$0 \leq -2(10) + 6 \Rightarrow 0 \leq -20 + 6 \Rightarrow 0 \leq -14$ (False)

$10 - 0 < 6 \Rightarrow 10 < 6$ (False)

Since both inequalities are false, $(10, 0)$ is not a solution.

3. Final Answer

The solutions are A and D.