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Sagot :
To solve and graph this system of inequalities, follow these steps:
### Step 1: Understand and Rewrite the Inequalities
1. The first line given is:
[tex]\[ y = \frac{1}{4} + 3 \][/tex]
This simplifies to:
[tex]\[ y = \frac{1}{4}x + 3 \][/tex]
2. The second inequality is:
[tex]\[ y \geq -5 \][/tex]
### Step 2: Plot the First Equation
We plot the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex]:
- Identify the y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ y = 3 \][/tex]
So the point (0, 3) is on the graph.
- Identify the slope, which is [tex]\(\frac{1}{4}\)[/tex]:
For every increase of 1 unit in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by [tex]\(\frac{1}{4}\)[/tex].
- Plot a few more points:
- When [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{4} \cdot 4 + 3 = 1 + 3 = 4 \)[/tex], so the point (4, 4) is on the graph.
- When [tex]\( x = -4 \)[/tex], [tex]\( y = \frac{1}{4} \cdot (-4) + 3 = -1 + 3 = 2 \)[/tex], so the point (-4, 2) is on the graph.
Connect these points with a straight line, extending it in both directions.
### Step 3: Represent the Inequality [tex]\( y \geq -5 \)[/tex]
To graph [tex]\( y \geq -5 \)[/tex]:
- Draw a horizontal line at [tex]\( y = -5 \)[/tex].
- Shade the region above this line including the line itself (because [tex]\( y \geq -5 \)[/tex] means all values of [tex]\( y \)[/tex] that are larger than or equal to -5).
### Step 4: Identify the Feasible Region
The feasible region is where the shaded area of [tex]\( y \geq -5 \)[/tex] overlaps with the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex]:
- Start from the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex].
- The graph of [tex]\( y \geq -5 \)[/tex] covers everything above the line [tex]\( y = -5 \)[/tex] up to continuing indefinitely upwards, while the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex] is a line extending diagonally.
### Text Description for the Graph
1. Line Plot: Draw the line from the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex] starting at the y-intercept (0, 3). As this is a linear function with a positive slope, the line will rise from left to right.
2. Horizontal Line and Shading: Draw a horizontal line at [tex]\( y = -5 \)[/tex] and shade the region above and including this line.
3. Overlap Region: Indicate that the feasible region (solution set) lies both above the [tex]\( y = -5 \)[/tex] line and on the [tex]\( y = \frac{1}{4}x + 3 \)[/tex] line.
This gives us the area where any point will satisfy both inequalities. The solution to the system is the region on and above the line [tex]\( y = -5 \)[/tex] that also lies on the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex].
### Step 1: Understand and Rewrite the Inequalities
1. The first line given is:
[tex]\[ y = \frac{1}{4} + 3 \][/tex]
This simplifies to:
[tex]\[ y = \frac{1}{4}x + 3 \][/tex]
2. The second inequality is:
[tex]\[ y \geq -5 \][/tex]
### Step 2: Plot the First Equation
We plot the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex]:
- Identify the y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ y = 3 \][/tex]
So the point (0, 3) is on the graph.
- Identify the slope, which is [tex]\(\frac{1}{4}\)[/tex]:
For every increase of 1 unit in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by [tex]\(\frac{1}{4}\)[/tex].
- Plot a few more points:
- When [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{4} \cdot 4 + 3 = 1 + 3 = 4 \)[/tex], so the point (4, 4) is on the graph.
- When [tex]\( x = -4 \)[/tex], [tex]\( y = \frac{1}{4} \cdot (-4) + 3 = -1 + 3 = 2 \)[/tex], so the point (-4, 2) is on the graph.
Connect these points with a straight line, extending it in both directions.
### Step 3: Represent the Inequality [tex]\( y \geq -5 \)[/tex]
To graph [tex]\( y \geq -5 \)[/tex]:
- Draw a horizontal line at [tex]\( y = -5 \)[/tex].
- Shade the region above this line including the line itself (because [tex]\( y \geq -5 \)[/tex] means all values of [tex]\( y \)[/tex] that are larger than or equal to -5).
### Step 4: Identify the Feasible Region
The feasible region is where the shaded area of [tex]\( y \geq -5 \)[/tex] overlaps with the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex]:
- Start from the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex].
- The graph of [tex]\( y \geq -5 \)[/tex] covers everything above the line [tex]\( y = -5 \)[/tex] up to continuing indefinitely upwards, while the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex] is a line extending diagonally.
### Text Description for the Graph
1. Line Plot: Draw the line from the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex] starting at the y-intercept (0, 3). As this is a linear function with a positive slope, the line will rise from left to right.
2. Horizontal Line and Shading: Draw a horizontal line at [tex]\( y = -5 \)[/tex] and shade the region above and including this line.
3. Overlap Region: Indicate that the feasible region (solution set) lies both above the [tex]\( y = -5 \)[/tex] line and on the [tex]\( y = \frac{1}{4}x + 3 \)[/tex] line.
This gives us the area where any point will satisfy both inequalities. The solution to the system is the region on and above the line [tex]\( y = -5 \)[/tex] that also lies on the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex].
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