At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
