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Sagot :
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].
[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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