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Sagot :
To solve the system of inequalities
[tex]\[ \begin{cases} -x + 2y \leq 4 \\ 3x + y > 1 \end{cases} \][/tex]
we need to determine the region in the [tex]\(xy\)[/tex]-plane that satisfies both inequalities. Let's break down each inequality and then find their intersection.
### Inequality 1: [tex]\(-x + 2y \leq 4\)[/tex]
1. Rewrite in slope-intercept form: Convert the inequality to the form [tex]\(y = mx + b\)[/tex].
[tex]\[ -x + 2y \leq 4 \implies 2y \leq x + 4 \implies y \leq \frac{1}{2}x + 2 \][/tex]
2. Graph the boundary line: Plot the line [tex]\(y = \frac{1}{2}x + 2\)[/tex]. This is a straight line with a slope of [tex]\(\frac{1}{2}\)[/tex] and a y-intercept at [tex]\((0, 2)\)[/tex]:
```
-1 | 5
0 | 4
1 | 3
2 | 2
3 | 1
4 | 0
5 | -1
```
3. Shading the region: Since the inequality is [tex]\(\leq\)[/tex], shade the region below and on the line [tex]\(y = \frac{1}{2}x + 2\)[/tex].
### Inequality 2: [tex]\(3x + y > 1\)[/tex]
1. Rewrite in slope-intercept form: Convert the inequality to the form [tex]\(y = mx + b\)[/tex].
[tex]\[ 3x + y > 1 \implies y > -3x + 1 \][/tex]
2. Graph the boundary line: Plot the line [tex]\(y = -3x + 1\)[/tex]. This is a straight line with a slope of [tex]\(-3\)[/tex] and a y-intercept at [tex]\((0, 1)\)[/tex]:
```
-1 | 4
0 | 1
1 | -2
2 | -5
```
3. Shading the region: Since the inequality is [tex]\(>\)[/tex], shade the region above the line [tex]\(y = -3x + 1\)[/tex].
### Finding the intersection
The solution to the system of inequalities is the region where the shaded areas of both inequalities overlap.
1. Find intersection points: Solve the boundary line equations simultaneously to find the points of intersection.
[tex]\[ \begin{cases} y = \frac{1}{2}x + 2 \\ y = -3x + 1 \end{cases} \][/tex]
Set the equations equal to each other to find the value of [tex]\(x\)[/tex]:
[tex]\[ \frac{1}{2}x + 2 = -3x + 1 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ \frac{1}{2}x + 3x = 1 - 2 \implies \frac{7}{2}x = -1 \implies x = -\frac{2}{7} \][/tex]
Substitute [tex]\(x = -\frac{2}{7}\)[/tex] back into one of the equations to find [tex]\(y\)[/tex]:
[tex]\[ y = -3\left(-\frac{2}{7}\right) + 1 = \frac{6}{7} + 1 = \frac{13}{7} \][/tex]
So, the intersection point is [tex]\(\left(-\frac{2}{7}, \frac{13}{7}\right)\)[/tex].
2. Graph the feasible region: The feasible region is the overlap of the shaded areas. It includes points below and on the line [tex]\(y \leq \frac{1}{2}x + 2\)[/tex] and above the line [tex]\(y > -3x + 1\)[/tex].
### Conclusion
The solution to the system of inequalities [tex]\(-x + 2y \leq 4\)[/tex] and [tex]\(3x + y > 1\)[/tex] is the region in the [tex]\(xy\)[/tex]-plane where both conditions are satisfied. This region lies below and on the line [tex]\(y = \frac{1}{2}x + 2\)[/tex] and above the line [tex]\(y = -3x + 1\)[/tex], and specifically in the area where these two shaded regions overlap.
[tex]\[ \begin{cases} -x + 2y \leq 4 \\ 3x + y > 1 \end{cases} \][/tex]
we need to determine the region in the [tex]\(xy\)[/tex]-plane that satisfies both inequalities. Let's break down each inequality and then find their intersection.
### Inequality 1: [tex]\(-x + 2y \leq 4\)[/tex]
1. Rewrite in slope-intercept form: Convert the inequality to the form [tex]\(y = mx + b\)[/tex].
[tex]\[ -x + 2y \leq 4 \implies 2y \leq x + 4 \implies y \leq \frac{1}{2}x + 2 \][/tex]
2. Graph the boundary line: Plot the line [tex]\(y = \frac{1}{2}x + 2\)[/tex]. This is a straight line with a slope of [tex]\(\frac{1}{2}\)[/tex] and a y-intercept at [tex]\((0, 2)\)[/tex]:
```
-1 | 5
0 | 4
1 | 3
2 | 2
3 | 1
4 | 0
5 | -1
```
3. Shading the region: Since the inequality is [tex]\(\leq\)[/tex], shade the region below and on the line [tex]\(y = \frac{1}{2}x + 2\)[/tex].
### Inequality 2: [tex]\(3x + y > 1\)[/tex]
1. Rewrite in slope-intercept form: Convert the inequality to the form [tex]\(y = mx + b\)[/tex].
[tex]\[ 3x + y > 1 \implies y > -3x + 1 \][/tex]
2. Graph the boundary line: Plot the line [tex]\(y = -3x + 1\)[/tex]. This is a straight line with a slope of [tex]\(-3\)[/tex] and a y-intercept at [tex]\((0, 1)\)[/tex]:
```
-1 | 4
0 | 1
1 | -2
2 | -5
```
3. Shading the region: Since the inequality is [tex]\(>\)[/tex], shade the region above the line [tex]\(y = -3x + 1\)[/tex].
### Finding the intersection
The solution to the system of inequalities is the region where the shaded areas of both inequalities overlap.
1. Find intersection points: Solve the boundary line equations simultaneously to find the points of intersection.
[tex]\[ \begin{cases} y = \frac{1}{2}x + 2 \\ y = -3x + 1 \end{cases} \][/tex]
Set the equations equal to each other to find the value of [tex]\(x\)[/tex]:
[tex]\[ \frac{1}{2}x + 2 = -3x + 1 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ \frac{1}{2}x + 3x = 1 - 2 \implies \frac{7}{2}x = -1 \implies x = -\frac{2}{7} \][/tex]
Substitute [tex]\(x = -\frac{2}{7}\)[/tex] back into one of the equations to find [tex]\(y\)[/tex]:
[tex]\[ y = -3\left(-\frac{2}{7}\right) + 1 = \frac{6}{7} + 1 = \frac{13}{7} \][/tex]
So, the intersection point is [tex]\(\left(-\frac{2}{7}, \frac{13}{7}\right)\)[/tex].
2. Graph the feasible region: The feasible region is the overlap of the shaded areas. It includes points below and on the line [tex]\(y \leq \frac{1}{2}x + 2\)[/tex] and above the line [tex]\(y > -3x + 1\)[/tex].
### Conclusion
The solution to the system of inequalities [tex]\(-x + 2y \leq 4\)[/tex] and [tex]\(3x + y > 1\)[/tex] is the region in the [tex]\(xy\)[/tex]-plane where both conditions are satisfied. This region lies below and on the line [tex]\(y = \frac{1}{2}x + 2\)[/tex] and above the line [tex]\(y = -3x + 1\)[/tex], and specifically in the area where these two shaded regions overlap.
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