To determine which points are solutions to the given system of inequalities, we need to check each point against all three inequalities. The system of inequalities is:
[tex]\[
\begin{array}{l}
y \geq x+3 \\
y < 2x \\
x > 6
\end{array}
\][/tex]
Let's evaluate each point.
Point A: [tex]\((8, 10)\)[/tex]
1. [tex]\(y \geq x + 3 \implies 10 \geq 8 + 3 \implies 10 \geq 11\)[/tex] (False)
2. [tex]\(y < 2x \implies 10 < 2 \times 8 \implies 10 < 16\)[/tex] (True)
3. [tex]\(x > 6 \implies 8 > 6\)[/tex] (True)
Since the first inequality is not satisfied, point [tex]\((8, 10)\)[/tex] is not a solution.
Point B: [tex]\((5, 10)\)[/tex]
1. [tex]\(y \geq x + 3 \implies 10 \geq 5 + 3 \implies 10 \geq 8\)[/tex] (True)
2. [tex]\(y < 2x \implies 10 < 2 \times 5 \implies 10 < 10\)[/tex] (False)
3. [tex]\(x > 6 \implies 5 > 6\)[/tex] (False)
Since the second and third inequalities are not satisfied, point [tex]\((5, 10)\)[/tex] is not a solution.
Point C: [tex]\((9, 12)\)[/tex]
1. [tex]\(y \geq x + 3 \implies 12 \geq 9 + 3 \implies 12 \geq 12\)[/tex] (True)
2. [tex]\(y < 2x \implies 12 < 2 \times 9 \implies 12 < 18\)[/tex] (True)
3. [tex]\(x > 6 \implies 9 > 6\)[/tex] (True)
Since all three inequalities are satisfied, point [tex]\((9, 12)\)[/tex] is a solution.
Point D: [tex]\((8, 12)\)[/tex]
1. [tex]\(y \geq x + 3 \implies 12 \geq 8 + 3 \implies 12 \geq 11\)[/tex] (True)
2. [tex]\(y < 2x \implies 12 < 2 \times 8 \implies 12 < 16\)[/tex] (True)
3. [tex]\(x > 6 \implies 8 > 6\)[/tex] (True)
Since all three inequalities are satisfied, point [tex]\((8, 12)\)[/tex] is a solution.
Point E: [tex]\((7, 10)\)[/tex]
1. [tex]\(y \geq x + 3 \implies 10 \geq 7 + 3 \implies 10 \geq 10\)[/tex] (True)
2. [tex]\(y < 2x \implies 10 < 2 \times 7 \implies 10 < 14\)[/tex] (True)
3. [tex]\(x > 6 \implies 7 > 6\)[/tex] (True)
Since all three inequalities are satisfied, point [tex]\((7, 10)\)[/tex] is a solution.
Point F: [tex]\((6, 11)\)[/tex]
1. [tex]\(y \geq x + 3 \implies 11 \geq 6 + 3 \implies 11 \geq 9\)[/tex] (True)
2. [tex]\(y < 2x \implies 11 < 2 \times 6 \implies 11 < 12\)[/tex] (True)
3. [tex]\(x > 6 \implies 6 > 6\)[/tex] (False)
Since the third inequality is not satisfied, point [tex]\((6, 11)\)[/tex] is not a solution.
Therefore, the points that satisfy all the given inequalities are:
[tex]\[
\boxed{(9, 12), (8, 12), (7, 10)}
\][/tex]
