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Sagot :
Certainly! Let's go through each part of the problem step-by-step.
### Part A: Description of the Graph
We have the following system of inequalities:
[tex]\[ \begin{array}{l} 2x + 3y \leq 15 \\ x + y \geq 3 \end{array} \][/tex]
1. Graphing the Inequalities:
- For the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
- The boundary line is [tex]\(2x + 3y = 15\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(3y = 15 \Rightarrow y = 5\)[/tex]. This gives the point [tex]\((0, 5)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x = 15 \Rightarrow x = 7.5\)[/tex]. This gives the point [tex]\((7.5, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\leq\)[/tex], shade the area below and including the line.
- For the inequality [tex]\(x + y \geq 3\)[/tex]:
- The boundary line is [tex]\(x + y = 3\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 3\)[/tex]. This gives the point [tex]\((0, 3)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 3\)[/tex]. This gives the point [tex]\((3, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\geq\)[/tex], shade the area above and including the line.
2. Solution Set:
- The solution set is the region where the shadings of the two inequalities overlap.
- The lines are solid because the inequalities are [tex]\(\leq\)[/tex] and [tex]\(\geq\)[/tex], which include the boundaries.
### Part B: Checking the Point (5,1)
Given the point [tex]\((5,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(5) + 3(1) = 10 + 3 = 13 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 5 + 1 = 6 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((5,1)\)[/tex] is included in the solution area.
### Part C: Choosing a Different Point
Let's choose the point [tex]\((3,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(3) + 3(1) = 6 + 3 = 9 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 3 + 1 = 4 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((3,1)\)[/tex] is in the solution set.
Interpretation:
- If Michael buys 3 cupcakes and 1 piece of fudge:
- The total cost would be [tex]\(2 \times 3 + 3 \times 1 = 6 + 3 = 9\)[/tex] dollars, which is within his budget of \$15.
- This combination would feed [tex]\(3 + 1 = 4\)[/tex] siblings, which meets the requirement of feeding at least 3 siblings.
Thus, the chosen point [tex]\((3,1)\)[/tex] provides a feasible and valid combination of cupcakes and fudge within the given constraints.
### Part A: Description of the Graph
We have the following system of inequalities:
[tex]\[ \begin{array}{l} 2x + 3y \leq 15 \\ x + y \geq 3 \end{array} \][/tex]
1. Graphing the Inequalities:
- For the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
- The boundary line is [tex]\(2x + 3y = 15\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(3y = 15 \Rightarrow y = 5\)[/tex]. This gives the point [tex]\((0, 5)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x = 15 \Rightarrow x = 7.5\)[/tex]. This gives the point [tex]\((7.5, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\leq\)[/tex], shade the area below and including the line.
- For the inequality [tex]\(x + y \geq 3\)[/tex]:
- The boundary line is [tex]\(x + y = 3\)[/tex].
- To graph this, find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 3\)[/tex]. This gives the point [tex]\((0, 3)\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 3\)[/tex]. This gives the point [tex]\((3, 0)\)[/tex].
- Plot these points and draw the line. Since the inequality is [tex]\(\geq\)[/tex], shade the area above and including the line.
2. Solution Set:
- The solution set is the region where the shadings of the two inequalities overlap.
- The lines are solid because the inequalities are [tex]\(\leq\)[/tex] and [tex]\(\geq\)[/tex], which include the boundaries.
### Part B: Checking the Point (5,1)
Given the point [tex]\((5,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(5) + 3(1) = 10 + 3 = 13 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 5 + 1 = 6 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((5,1)\)[/tex] is included in the solution area.
### Part C: Choosing a Different Point
Let's choose the point [tex]\((3,1)\)[/tex]:
1. Check the inequality [tex]\(2x + 3y \leq 15\)[/tex]:
[tex]\[ 2(3) + 3(1) = 6 + 3 = 9 \leq 15 \][/tex]
This condition is satisfied.
2. Check the inequality [tex]\(x + y \geq 3\)[/tex]:
[tex]\[ 3 + 1 = 4 \geq 3 \][/tex]
This condition is also satisfied.
Since both conditions are satisfied, the point [tex]\((3,1)\)[/tex] is in the solution set.
Interpretation:
- If Michael buys 3 cupcakes and 1 piece of fudge:
- The total cost would be [tex]\(2 \times 3 + 3 \times 1 = 6 + 3 = 9\)[/tex] dollars, which is within his budget of \$15.
- This combination would feed [tex]\(3 + 1 = 4\)[/tex] siblings, which meets the requirement of feeding at least 3 siblings.
Thus, the chosen point [tex]\((3,1)\)[/tex] provides a feasible and valid combination of cupcakes and fudge within the given constraints.
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