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Which ordered pair is the solution to the system of equations?

[tex]\[
\left\{
\begin{array}{l}
x + 2y = -8 \\
y = \frac{1}{2}x - 2
\end{array}
\right.
\][/tex]

A. [tex]\((-2, -1)\)[/tex]
B. [tex]\((-2, -3)\)[/tex]
C. [tex]\((-6, -5)\)[/tex]
D. [tex]\((-4, -4)\)[/tex]


Sagot :

To determine which ordered pair is the solution to the system of equations
[tex]\[ \left\{\begin{array}{l} x + 2y = -8 \\ y = \frac{1}{2}x - 2 \end{array}\right., \][/tex]
we need to verify which pair satisfies both equations.

Let's analyze each pair one by one.

### Pair [tex]\((-2, -1)\)[/tex]

1. Substitute [tex]\(x = -2\)[/tex] and [tex]\(y = -1\)[/tex] into the first equation [tex]\(x + 2y = -8\)[/tex]:
[tex]\[ -2 + 2(-1) = -2 - 2 = -4 \quad (\text{not } -8) \][/tex]
Hence, this pair does not satisfy the first equation.

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

1. Substitute [tex]\(x = -2\)[/tex] and [tex]\(y = -3\)[/tex] into the first equation [tex]\(x + 2y = -8\)[/tex]:
[tex]\[ -2 + 2(-3) = -2 - 6 = -8 \quad (\text{correct}) \][/tex]
2. Substitute [tex]\(x = -2\)[/tex] and [tex]\(y = -3\)[/tex] into the second equation [tex]\(y = \frac{1}{2}x - 2\)[/tex]:
[tex]\[ -3 = \frac{1}{2}(-2) - 2 = -1 - 2 = -3 \quad (\text{correct}) \][/tex]
This pair satisfies both equations.

### Pair [tex]\((-6, -5)\)[/tex]

1. Substitute [tex]\(x = -6\)[/tex] and [tex]\(y = -5\)[/tex] into the first equation [tex]\(x + 2y = -8\)[/tex]:
[tex]\[ -6 + 2(-5) = -6 - 10 = -16 \quad (\text{not } -8) \][/tex]
Hence, this pair does not satisfy the first equation.

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

1. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = -4\)[/tex] into the first equation [tex]\(x + 2y = -8\)[/tex]:
[tex]\[ -4 + 2(-4) = -4 - 8 = -12 \quad (\text{not } -8) \][/tex]
Hence, this pair does not satisfy the first equation.

Based on the above evaluations, the only ordered pair that satisfies both equations is:
[tex]\[ (-2, -3) \][/tex]

Thus, the solution to the system of equations is [tex]\(\boxed{(-2, -3)}\)[/tex].