Answered

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The following inequalities form a system:
[tex]\[
\begin{array}{l}
y \geq \frac{2}{3} x+1 \\
y\ \textless \ -\frac{1}{4} x+2
\end{array}
\][/tex]

Which ordered pair is included in the solution to this system?

A. [tex]$(-6, 3.5)$[/tex]
B. [tex]$(-6, -3)$[/tex]
C. [tex]$(-4, 3)$[/tex]
D. [tex]$(-4, 4)$[/tex]

Sagot :

GinnyAnswer
To determine which ordered pair is included in the solution to the following system of inequalities:
[tex]\[ \begin{array}{l} y \geq \frac{2}{3} x+1 \\ y < -\frac{1}{4} x + 2 \end{array} \][/tex]

we will need to check each ordered pair [tex]\((x, y)\)[/tex] to see if both inequalities are satisfied.

### Checking [tex]\((-6, 3.5)\)[/tex]:
1. Substitute [tex]\(x = -6\)[/tex] and [tex]\(y = 3.5\)[/tex] into the first inequality [tex]\(y \geq \frac{2}{3} x + 1\)[/tex]:
[tex]\[ 3.5 \geq \frac{2}{3}(-6) + 1 \implies 3.5 \geq -4 + 1 \implies 3.5 \geq -3 \][/tex]
This is true.

2. Substitute [tex]\(x = -6\)[/tex] and [tex]\(y = 3.5\)[/tex] into the second inequality [tex]\(y < -\frac{1}{4} x + 2\)[/tex]:
[tex]\[ 3.5 < -\frac{1}{4}(-6) + 2 \implies 3.5 < 1.5 + 2 \implies 3.5 < 3.5 \][/tex]
This is not true (3.5 is not less than 3.5).

Since the second inequality is not satisfied, [tex]\((-6, 3.5)\)[/tex] is not a solution.

### Checking [tex]\((-6, -3)\)[/tex]:
1. Substitute [tex]\(x = -6\)[/tex] and [tex]\(y = -3\)[/tex] into the first inequality [tex]\(y \geq \frac{2}{3} x + 1\)[/tex]:
[tex]\[ -3 \geq \frac{2}{3}(-6) + 1 \implies -3 \geq -4 + 1 \implies -3 \geq -3 \][/tex]
This is true.

2. Substitute [tex]\(x = -6\)[/tex] and [tex]\(y = -3\)[/tex] into the second inequality [tex]\(y < -\frac{1}{4} x + 2\)[/tex]:
[tex]\[ -3 < -\frac{1}{4}(-6) + 2 \implies -3 < 1.5 + 2 \implies -3 < 3.5 \][/tex]
This is true.

Since both inequalities are satisfied, [tex]\((-6, -3)\)[/tex] is a solution.

### Checking [tex]\((-4, 3)\)[/tex]:
1. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 3\)[/tex] into the first inequality [tex]\(y \geq \frac{2}{3} x + 1\)[/tex]:
[tex]\[ 3 \geq \frac{2}{3}(-4) + 1 \implies 3 \geq -\frac{8}{3} + 1 \implies 3 \geq -\frac{5}{3} \][/tex]
This is true.

2. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 3\)[/tex] into the second inequality [tex]\(y < -\frac{1}{4} x + 2\)[/tex]:
[tex]\[ 3 < -\frac{1}{4}(-4) + 2 \implies 3 < 1 + 2 \implies 3 < 3 \][/tex]
This is not true.

Since the second inequality is not satisfied, [tex]\((-4, 3)\)[/tex] is not a solution.

### Checking [tex]\((-4, 4)\)[/tex]:
1. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 4\)[/tex] into the first inequality [tex]\(y \geq \frac{2}{3} x + 1\)[/tex]:
[tex]\[ 4 \geq \frac{2}{3}(-4) + 1 \implies 4 \geq -\frac{8}{3} + 1 \implies 4 \geq -\frac{5}{3} \][/tex]
This is true.

2. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 4\)[/tex] into the second inequality [tex]\(y < -\frac{1}{4} x + 2\)[/tex]:
[tex]\[ 4 < -\frac{1}{4}(-4) + 2 \implies 4 < 1 + 2 \implies 4 < 3 \][/tex]
This is not true.

Since the second inequality is not satisfied, [tex]\((-4, 4)\)[/tex] is not a solution.

Therefore, the only ordered pair that satisfies both inequalities in the system is [tex]\((-6, -3)\)[/tex].
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