Which ordered pair is a solution to the following system of inequalities?

[tex]\[
\begin{array}{l}
y \leq -x^2 + 8x \\
y \ \textgreater \ x^2 - 3
\end{array}
\][/tex]

A. [tex]$(-1, 4)$[/tex]
B. [tex]$(0, 9)$[/tex]
C. [tex]$(2, 7)$[/tex]
D. [tex]$(4, 3)$[/tex]



Answer :

To determine which ordered pair is a solution to the system of inequalities:

[tex]\[ \begin{cases} y \leq -x^2 + 8x \\ y > x^2 - 3 \end{cases} \][/tex]

We must check each pair to see if it satisfies both inequalities.

### Pair 1: [tex]\((-1, 4)\)[/tex]

1. Check [tex]\( y \leq -x^2 + 8x \)[/tex]:
[tex]\( y = 4 \)[/tex]
[tex]\( x = -1 \)[/tex]
[tex]\[ -(-1)^2 + 8(-1) = -1 - 8 = -9 \][/tex]
[tex]\[ 4 \leq -9 \quad \text{(False)} \][/tex]

### Pair 2: [tex]\((0, 9)\)[/tex]

1. Check [tex]\( y \leq -x^2 + 8x \)[/tex]:
[tex]\( y = 9 \)[/tex]
[tex]\( x = 0 \)[/tex]
[tex]\[ -0^2 + 8(0) = 0 \][/tex]
[tex]\[ 9 \leq 0 \quad \text{(False)} \][/tex]

### Pair 3: [tex]\((2, 7)\)[/tex]

1. Check [tex]\( y \leq -x^2 + 8x \)[/tex]:
[tex]\( y = 7 \)[/tex]
[tex]\( x = 2 \)[/tex]
[tex]\[ -(2)^2 + 8(2) = -4 + 16 = 12 \][/tex]
[tex]\[ 7 \leq 12 \quad \text{(True)} \][/tex]

2. Check [tex]\( y > x^2 - 3 \)[/tex]:
[tex]\( y = 7 \)[/tex]
[tex]\[ (2)^2 - 3 = 4 - 3 = 1 \][/tex]
[tex]\[ 7 > 1 \quad \text{(True)} \][/tex]

Since both conditions are satisfied, [tex]\((2, 7)\)[/tex] is a solution.

### Pair 4: [tex]\((4, 3)\)[/tex]

1. Check [tex]\( y \leq -x^2 + 8x \)[/tex]:
[tex]\( y = 3 \)[/tex]
[tex]\( x = 4 \)[/tex]
[tex]\[ -(4)^2 + 8(4) = -16 + 32 = 16 \][/tex]
[tex]\[ 3 \leq 16 \quad \text{(True)} \][/tex]

2. Check [tex]\( y > x^2 - 3 \)[/tex]:
[tex]\( y = 3 \)[/tex]
[tex]\[ (4)^2 - 3 = 16 - 3 = 13 \][/tex]
[tex]\[ 3 > 13 \quad \text{(False)} \][/tex]

Since [tex]\((4, 3)\)[/tex] does not satisfy the second inequality, it is not a solution.

Thus, the ordered pair that satisfies both inequalities is [tex]\((2, 7)\)[/tex].

The correct answer is:
[tex]\[ (2, 7) \][/tex]