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Sagot :
To graph the solution to the system of inequalities:
[tex]\[ \begin{array}{l} 2x - 3y < 15 \\ y \leq x + 2 \end{array} \][/tex]
we need to take the following step-by-step approach:
### Step 1: Graph the boundary lines
1. Equation for the first inequality: \(2x - 3y = 15\)
- To graph this, find two points on the line:
- Set \(x = 0\): \(2(0) - 3y = 15 \implies -3y = 15 \implies y = -5\). So, one point is \((0, -5)\).
- Set \(y = 0\): \(2x - 3(0) = 15 \implies 2x = 15 \implies x = 7.5\). So, another point is \((7.5, 0)\).
- Draw a dashed line through these points since the inequality is strictly less than \( (<) \).
2. Equation for the second inequality: \(y = x + 2\)
- To graph this, find two points on the line:
- Set \(x = 0\): \(y = 0 + 2 = 2\). So, one point is \((0, 2)\).
- Set \(x = -2\): \(y = -2 + 2 = 0\). So, another point is \((-2, 0)\).
- Draw a solid line through these points since the inequality is less than or equal to \( (≤) \).
### Step 2: Determine the shading for each inequality
1. Shading for the first inequality: \(2x - 3y < 15\)
- Choose a test point not on the line, like \((0, 0)\):
- Plug it into the inequality: \(2(0) - 3(0) < 15 \implies 0 < 15\) (True).
- Since the inequality holds true, shade the region below the line \(2x - 3y = 15\).
2. Shading for the second inequality: \(y \leq x + 2\)
- Choose a test point not on the line, like \((0, 0)\):
- Plug it into the inequality: \(0 \leq 0 + 2 \implies 0 \leq 2\) (True).
- Since the inequality holds true, shade the region below the line \(y = x + 2\).
### Step 3: Find the intersection of the shaded regions
The solution to the system of inequalities is the overlapping shaded region.
### Graph the Solution
1. Draw the dashed line for \(2x - 3y = 15\).
2. Draw the solid line for \(y = x + 2\).
3. Shade the region below the line \(2x - 3y = 15\).
4. Shade the region below the line \(y = x + 2\).
The overlapping shaded area represents the solution to the system of inequalities.
Here is a description for what the graph should look like:
- Two lines intersecting.
- The dashed line for \(2x - 3y = 15\) indicating values below it are valid.
- The solid line for \(y = x + 2\).
- The region below both lines will be shaded, indicating the set of points that satisfy both inequalities.
Use graph paper or a coordinate plane tool to illustrate this and ensure correct shading for clarity.
[tex]\[ \begin{array}{l} 2x - 3y < 15 \\ y \leq x + 2 \end{array} \][/tex]
we need to take the following step-by-step approach:
### Step 1: Graph the boundary lines
1. Equation for the first inequality: \(2x - 3y = 15\)
- To graph this, find two points on the line:
- Set \(x = 0\): \(2(0) - 3y = 15 \implies -3y = 15 \implies y = -5\). So, one point is \((0, -5)\).
- Set \(y = 0\): \(2x - 3(0) = 15 \implies 2x = 15 \implies x = 7.5\). So, another point is \((7.5, 0)\).
- Draw a dashed line through these points since the inequality is strictly less than \( (<) \).
2. Equation for the second inequality: \(y = x + 2\)
- To graph this, find two points on the line:
- Set \(x = 0\): \(y = 0 + 2 = 2\). So, one point is \((0, 2)\).
- Set \(x = -2\): \(y = -2 + 2 = 0\). So, another point is \((-2, 0)\).
- Draw a solid line through these points since the inequality is less than or equal to \( (≤) \).
### Step 2: Determine the shading for each inequality
1. Shading for the first inequality: \(2x - 3y < 15\)
- Choose a test point not on the line, like \((0, 0)\):
- Plug it into the inequality: \(2(0) - 3(0) < 15 \implies 0 < 15\) (True).
- Since the inequality holds true, shade the region below the line \(2x - 3y = 15\).
2. Shading for the second inequality: \(y \leq x + 2\)
- Choose a test point not on the line, like \((0, 0)\):
- Plug it into the inequality: \(0 \leq 0 + 2 \implies 0 \leq 2\) (True).
- Since the inequality holds true, shade the region below the line \(y = x + 2\).
### Step 3: Find the intersection of the shaded regions
The solution to the system of inequalities is the overlapping shaded region.
### Graph the Solution
1. Draw the dashed line for \(2x - 3y = 15\).
2. Draw the solid line for \(y = x + 2\).
3. Shade the region below the line \(2x - 3y = 15\).
4. Shade the region below the line \(y = x + 2\).
The overlapping shaded area represents the solution to the system of inequalities.
Here is a description for what the graph should look like:
- Two lines intersecting.
- The dashed line for \(2x - 3y = 15\) indicating values below it are valid.
- The solid line for \(y = x + 2\).
- The region below both lines will be shaded, indicating the set of points that satisfy both inequalities.
Use graph paper or a coordinate plane tool to illustrate this and ensure correct shading for clarity.
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