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Identify any solutions to the system shown here.

[tex]\[
\begin{array}{l}
2x + 3y \geq 6 \\
3x + 2y \leq 6
\end{array}
\][/tex]

A. [tex]\((1.5, 1)\)[/tex]

B. [tex]\((0.5, 2)\)[/tex]

C. [tex]\((-1, 2.5)\)[/tex]

D. [tex]\((-2, 4)\)[/tex]


Sagot :

To identify solutions to the given system of inequalities, we will test each point by substituting the coordinates [tex]\( (x, y) \)[/tex] into the inequalities and verifying if they satisfy both conditions simultaneously. The system of inequalities is:

[tex]\[ \begin{cases} 2x + 3y \geq 6 \\ 3x + 2y \leq 6 \end{cases} \][/tex]

Let's check the given points one by one:

1. Point [tex]\( (1.5, 1) \)[/tex]:
- For [tex]\( 2x + 3y \geq 6 \)[/tex]:
[tex]\[ 2(1.5) + 3(1) = 3 + 3 = 6 \quad \text{(True)} \][/tex]
- For [tex]\( 3x + 2y \leq 6 \)[/tex]:
[tex]\[ 3(1.5) + 2(1) = 4.5 + 2 = 6.5 \quad \text{(False)} \][/tex]
Since the second inequality is not satisfied, [tex]\( (1.5, 1) \)[/tex] is not a solution.

2. Point [tex]\( (0.5, 2) \)[/tex]:
- For [tex]\( 2x + 3y \geq 6 \)[/tex]:
[tex]\[ 2(0.5) + 3(2) = 1 + 6 = 7 \quad \text{(True)} \][/tex]
- For [tex]\( 3x + 2y \leq 6 \)[/tex]:
[tex]\[ 3(0.5) + 2(2) = 1.5 + 4 = 5.5 \quad \text{(True)} \][/tex]
Since both inequalities are satisfied, [tex]\( (0.5, 2) \)[/tex] is a solution.

3. Point [tex]\( (-1, 2.5) \)[/tex]:
- For [tex]\( 2x + 3y \geq 6 \)[/tex]:
[tex]\[ 2(-1) + 3(2.5) = -2 + 7.5 = 5.5 \quad \text{(False)} \][/tex]
- For [tex]\( 3x + 2y \leq 6 \)[/tex]:
[tex]\[ 3(-1) + 2(2.5) = -3 + 5 = 2 \quad \text{(True)} \][/tex]
Since the first inequality is not satisfied, [tex]\( (-1, 2.5) \)[/tex] is not a solution.

4. Point [tex]\( (-2, 4) \)[/tex]:
- For [tex]\( 2x + 3y \geq 6 \)[/tex]:
[tex]\[ 2(-2) + 3(4) = -4 + 12 = 8 \quad \text{(True)} \][/tex]
- For [tex]\( 3x + 2y \leq 6 \)[/tex]:
[tex]\[ 3(-2) + 2(4) = -6 + 8 = 2 \quad \text{(True)} \][/tex]
Since both inequalities are satisfied, [tex]\( (-2, 4) \)[/tex] is a solution.

Thus, the points that are solutions to the system of inequalities are:
[tex]\[ (0.5, 2) \quad \text{and} \quad (-2, 4) \][/tex]