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Sagot :
Sure, I'll guide you through the process of graphing the system of inequalities:
1. Inequality 1: [tex]\( x + 3y > -3 \)[/tex]
- First, let's convert the inequality to an equation to plot: [tex]\( x + 3y = -3 \)[/tex].
- To find the intercepts:
- For the [tex]\( y \)[/tex]-intercept, set [tex]\( x = 0 \)[/tex]:
[tex]\[ 0 + 3y = -3 \implies y = -1 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, -1) \)[/tex].
- For the [tex]\( x \)[/tex]-intercept, set [tex]\( y = 0 \)[/tex]:
[tex]\[ x + 0 = -3 \implies x = -3 \][/tex]
So, the [tex]\( x \)[/tex]-intercept is [tex]\( (-3, 0) \)[/tex].
- Now plot the line passing through [tex]\( (0, -1) \)[/tex] and [tex]\( (-3, 0) \)[/tex]. This will be a dashed line because the inequality is a strict inequality ([tex]\( > \)[/tex]).
- The inequality [tex]\( x + 3y > -3 \)[/tex] means we shade the region above the line. This is because for any point [tex]\( (x, y) \)[/tex] above the line, the sum [tex]\( x + 3y \)[/tex] will be greater than [tex]\(-3\)[/tex].
2. Inequality 2: [tex]\( y < \frac{1}{2} x + 1 \)[/tex]
- Convert to an equation to plot: [tex]\( y = \frac{1}{2} x + 1 \)[/tex].
- This is a line with slope [tex]\( \frac{1}{2} \)[/tex] and [tex]\( y \)[/tex]-intercept [tex]\( 1 \)[/tex].
- To find another point on the line:
- If [tex]\( x = 2 \)[/tex], then:
[tex]\[ y = \frac{1}{2}(2) + 1 = 1 + 1 = 2 \][/tex]
So, another point is [tex]\( (2, 2) \)[/tex].
- Now plot the line passing through [tex]\( (0, 1) \)[/tex] and [tex]\( (2, 2) \)[/tex]. This will also be a dashed line because the inequality is strict ([tex]\( < \)[/tex]).
- The inequality [tex]\( y < \frac{1}{2} x + 1 \)[/tex] means we shade the region below the line. This is because for any point [tex]\( (x, y) \)[/tex] below the line, [tex]\( y \)[/tex] will be less than [tex]\(\frac{1}{2} x + 1 \)[/tex].
3. Graphing the System
- Combine both shaded regions from the two inequalities on the same graph.
- The feasible region, which is the solution to the system, will be where both shaded regions overlap.
4. Summary
- Plot the line [tex]\( x + 3y = -3 \)[/tex] as a dashed line.
- Shade the region above this line.
- Plot the line [tex]\( y = \frac{1}{2} x + 1 \)[/tex] as a dashed line.
- Shade the region below this line.
- The overlapping shaded region is the solution to the system [tex]\( x + 3y > -3 \)[/tex] and [tex]\( y < \frac{1}{2} x + 1 \)[/tex].
By following these steps, you’ll have the graph of the system of inequalities plotted correctly.
1. Inequality 1: [tex]\( x + 3y > -3 \)[/tex]
- First, let's convert the inequality to an equation to plot: [tex]\( x + 3y = -3 \)[/tex].
- To find the intercepts:
- For the [tex]\( y \)[/tex]-intercept, set [tex]\( x = 0 \)[/tex]:
[tex]\[ 0 + 3y = -3 \implies y = -1 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, -1) \)[/tex].
- For the [tex]\( x \)[/tex]-intercept, set [tex]\( y = 0 \)[/tex]:
[tex]\[ x + 0 = -3 \implies x = -3 \][/tex]
So, the [tex]\( x \)[/tex]-intercept is [tex]\( (-3, 0) \)[/tex].
- Now plot the line passing through [tex]\( (0, -1) \)[/tex] and [tex]\( (-3, 0) \)[/tex]. This will be a dashed line because the inequality is a strict inequality ([tex]\( > \)[/tex]).
- The inequality [tex]\( x + 3y > -3 \)[/tex] means we shade the region above the line. This is because for any point [tex]\( (x, y) \)[/tex] above the line, the sum [tex]\( x + 3y \)[/tex] will be greater than [tex]\(-3\)[/tex].
2. Inequality 2: [tex]\( y < \frac{1}{2} x + 1 \)[/tex]
- Convert to an equation to plot: [tex]\( y = \frac{1}{2} x + 1 \)[/tex].
- This is a line with slope [tex]\( \frac{1}{2} \)[/tex] and [tex]\( y \)[/tex]-intercept [tex]\( 1 \)[/tex].
- To find another point on the line:
- If [tex]\( x = 2 \)[/tex], then:
[tex]\[ y = \frac{1}{2}(2) + 1 = 1 + 1 = 2 \][/tex]
So, another point is [tex]\( (2, 2) \)[/tex].
- Now plot the line passing through [tex]\( (0, 1) \)[/tex] and [tex]\( (2, 2) \)[/tex]. This will also be a dashed line because the inequality is strict ([tex]\( < \)[/tex]).
- The inequality [tex]\( y < \frac{1}{2} x + 1 \)[/tex] means we shade the region below the line. This is because for any point [tex]\( (x, y) \)[/tex] below the line, [tex]\( y \)[/tex] will be less than [tex]\(\frac{1}{2} x + 1 \)[/tex].
3. Graphing the System
- Combine both shaded regions from the two inequalities on the same graph.
- The feasible region, which is the solution to the system, will be where both shaded regions overlap.
4. Summary
- Plot the line [tex]\( x + 3y = -3 \)[/tex] as a dashed line.
- Shade the region above this line.
- Plot the line [tex]\( y = \frac{1}{2} x + 1 \)[/tex] as a dashed line.
- Shade the region below this line.
- The overlapping shaded region is the solution to the system [tex]\( x + 3y > -3 \)[/tex] and [tex]\( y < \frac{1}{2} x + 1 \)[/tex].
By following these steps, you’ll have the graph of the system of inequalities plotted correctly.
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