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To determine which of the given points is a solution to the system of inequalities
[tex]\[ \begin{array}{l} y \leq 2x + 2 \\ y \geq -5x + 4 \end{array} \][/tex]
we will evaluate each point against both inequalities to check if it satisfies them.
Let's analyze each point one by one:
### Point A: [tex]\((1, 6)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 6\quad \leq \quad 2(1) + 2 = 4 \][/tex]
This is false, since [tex]\(6 \leq 4\)[/tex] is not true.
Since one of the inequalities is not satisfied, point [tex]\((1, 6)\)[/tex] is not a solution.
### Point B: [tex]\((-6, 0)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 0\quad \leq \quad 2(-6) + 2 = -12 + 2 = -10 \][/tex]
This is false, since [tex]\(0 \leq -10\)[/tex] is not true.
Again, since one of the inequalities is not satisfied, point [tex]\((-6, 0)\)[/tex] is not a solution.
### Point C: [tex]\((0, 5)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 5\quad \leq \quad 2(0) + 2 = 2 \][/tex]
This is false, since [tex]\(5 \leq 2\)[/tex] is not true.
Since one of the inequalities is not satisfied, point [tex]\((0, 5)\)[/tex] is not a solution.
### Point D: [tex]\((5, 0)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 0\quad \leq \quad 2(5) + 2 = 10 + 2 = 12 \][/tex]
This is true, since [tex]\(0 \leq 12\)[/tex].
2. For [tex]\( y \geq -5x + 4 \)[/tex]:
[tex]\[ y = 0\quad \geq \quad -5(5) + 4 = -25 + 4 = -21 \][/tex]
This is also true, since [tex]\(0 \geq -21\)[/tex].
Since point [tex]\((5, 0)\)[/tex] satisfies both inequalities, it is a solution.
### Conclusion
Therefore, the point that is a solution to the system of inequalities is:
[tex]\[ \boxed{(5, 0)} \][/tex]
[tex]\[ \begin{array}{l} y \leq 2x + 2 \\ y \geq -5x + 4 \end{array} \][/tex]
we will evaluate each point against both inequalities to check if it satisfies them.
Let's analyze each point one by one:
### Point A: [tex]\((1, 6)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 6\quad \leq \quad 2(1) + 2 = 4 \][/tex]
This is false, since [tex]\(6 \leq 4\)[/tex] is not true.
Since one of the inequalities is not satisfied, point [tex]\((1, 6)\)[/tex] is not a solution.
### Point B: [tex]\((-6, 0)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 0\quad \leq \quad 2(-6) + 2 = -12 + 2 = -10 \][/tex]
This is false, since [tex]\(0 \leq -10\)[/tex] is not true.
Again, since one of the inequalities is not satisfied, point [tex]\((-6, 0)\)[/tex] is not a solution.
### Point C: [tex]\((0, 5)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 5\quad \leq \quad 2(0) + 2 = 2 \][/tex]
This is false, since [tex]\(5 \leq 2\)[/tex] is not true.
Since one of the inequalities is not satisfied, point [tex]\((0, 5)\)[/tex] is not a solution.
### Point D: [tex]\((5, 0)\)[/tex]
1. For [tex]\( y \leq 2x + 2 \)[/tex]:
[tex]\[ y = 0\quad \leq \quad 2(5) + 2 = 10 + 2 = 12 \][/tex]
This is true, since [tex]\(0 \leq 12\)[/tex].
2. For [tex]\( y \geq -5x + 4 \)[/tex]:
[tex]\[ y = 0\quad \geq \quad -5(5) + 4 = -25 + 4 = -21 \][/tex]
This is also true, since [tex]\(0 \geq -21\)[/tex].
Since point [tex]\((5, 0)\)[/tex] satisfies both inequalities, it is a solution.
### Conclusion
Therefore, the point that is a solution to the system of inequalities is:
[tex]\[ \boxed{(5, 0)} \][/tex]
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