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To solve the system of inequalities:
[tex]\[ \begin{cases} x + 4y \geq 10 \\ 3x - 2y < 12 \end{cases} \][/tex]
we need to determine the region in the [tex]\(xy\)[/tex]-plane that satisfies both inequalities. We'll proceed step-by-step by analyzing each inequality and then finding the intersection of the regions they define.
### Analyzing the First Inequality: [tex]\(x + 4y \geq 10\)[/tex]
1. Rewrite in Slope-Intercept Form:
[tex]\[ x + 4y = 10 \quad \Rightarrow \quad 4y = -x + 10 \quad \Rightarrow \quad y = -\frac{1}{4}x + \frac{10}{4} \quad \Rightarrow \quad y = -\frac{1}{4}x + 2.5 \][/tex]
2. Graph the Equality:
- This line has a y-intercept of [tex]\(2.5\)[/tex] and a slope of [tex]\(-\frac{1}{4}\)[/tex].
- The line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex] divides the plane.
3. Determine the Region:
- To figure out which side of the line is valid, pick a test point such as (0, 0).
- Substitute into the inequality: [tex]\(0 + 4(0) = 0\)[/tex], we see that [tex]\(0 \geq 10\)[/tex] is false.
- So, the region that satisfies [tex]\(x + 4y \geq 10\)[/tex] is the region above the line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex].
### Analyzing the Second Inequality: [tex]\(3x - 2y < 12\)[/tex]
1. Rewrite in Slope-Intercept Form:
[tex]\[ 3x - 2y = 12 \quad \Rightarrow \quad -2y = -3x + 12 \quad \Rightarrow \quad y = \frac{3}{2}x - 6 \][/tex]
2. Graph the Equality:
- This line has a y-intercept of [tex]\(-6\)[/tex] and a slope of [tex]\(\frac{3}{2}\)[/tex].
- The line [tex]\(y = \frac{3}{2}x - 6\)[/tex] divides the plane.
3. Determine the Region:
- To figure out which side of the line is valid, pick a test point such as [tex]\((0, 0)\)[/tex].
- Substitute into the inequality: [tex]\(3(0) - 2(0) = 0\)[/tex], we see that [tex]\(0 < 12\)[/tex] is true.
- So, the region that satisfies [tex]\(3x - 2y < 12\)[/tex] is the region below the line [tex]\(y = \frac{3}{2}x - 6\)[/tex].
### Finding the Intersection of the Regions
We now seek the overlapping region that satisfies both inequalities:
1. Graphical Solution:
- Draw the lines [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex] and [tex]\(y = \frac{3}{2}x - 6\)[/tex] on the same graph.
- Identify the region that is both above the line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex] and below the line [tex]\(y = \frac{3}{2}x - 6\)[/tex].
2. Points of Intersection:
- Find where the lines intersect by solving the system:
[tex]\[ -\frac{1}{4}x + 2.5 = \frac{3}{2}x - 6 \][/tex]
Multiply through by 4 to clear fractions:
[tex]\[ -x + 10 = 6x - 24 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 10 + 24 = 7x \quad \Rightarrow \quad 34 = 7x \quad \Rightarrow \quad x = \frac{34}{7} \approx 4.857 \][/tex]
Substitute [tex]\(x = \frac{34}{7}\)[/tex] back into either line equation to find [tex]\(y\)[/tex]:
[tex]\[ y = -\frac{1}{4}\left(\frac{34}{7}\right) + 2.5 = \frac{-34}{28} + 2.5 = -\frac{17}{14} + 2.5 = -\frac{17}{14} + \frac{35}{14} = \frac{18}{14} = \frac{9}{7} \approx 1.286 \][/tex]
Therefore, the lines intersect at approximately [tex]\(\left(\frac{34}{7}, \frac{9}{7}\right)\)[/tex].
### Conclusion
The solution to the system of inequalities is the region in the [tex]\(xy\)[/tex]-plane that:
- Lies above the line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex]
- Lies below the line [tex]\(y = \frac{3}{2}x - 6\)[/tex]
Graphing these regions and finding their overlap will visually provide the area satisfying both conditions.
[tex]\[ \begin{cases} x + 4y \geq 10 \\ 3x - 2y < 12 \end{cases} \][/tex]
we need to determine the region in the [tex]\(xy\)[/tex]-plane that satisfies both inequalities. We'll proceed step-by-step by analyzing each inequality and then finding the intersection of the regions they define.
### Analyzing the First Inequality: [tex]\(x + 4y \geq 10\)[/tex]
1. Rewrite in Slope-Intercept Form:
[tex]\[ x + 4y = 10 \quad \Rightarrow \quad 4y = -x + 10 \quad \Rightarrow \quad y = -\frac{1}{4}x + \frac{10}{4} \quad \Rightarrow \quad y = -\frac{1}{4}x + 2.5 \][/tex]
2. Graph the Equality:
- This line has a y-intercept of [tex]\(2.5\)[/tex] and a slope of [tex]\(-\frac{1}{4}\)[/tex].
- The line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex] divides the plane.
3. Determine the Region:
- To figure out which side of the line is valid, pick a test point such as (0, 0).
- Substitute into the inequality: [tex]\(0 + 4(0) = 0\)[/tex], we see that [tex]\(0 \geq 10\)[/tex] is false.
- So, the region that satisfies [tex]\(x + 4y \geq 10\)[/tex] is the region above the line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex].
### Analyzing the Second Inequality: [tex]\(3x - 2y < 12\)[/tex]
1. Rewrite in Slope-Intercept Form:
[tex]\[ 3x - 2y = 12 \quad \Rightarrow \quad -2y = -3x + 12 \quad \Rightarrow \quad y = \frac{3}{2}x - 6 \][/tex]
2. Graph the Equality:
- This line has a y-intercept of [tex]\(-6\)[/tex] and a slope of [tex]\(\frac{3}{2}\)[/tex].
- The line [tex]\(y = \frac{3}{2}x - 6\)[/tex] divides the plane.
3. Determine the Region:
- To figure out which side of the line is valid, pick a test point such as [tex]\((0, 0)\)[/tex].
- Substitute into the inequality: [tex]\(3(0) - 2(0) = 0\)[/tex], we see that [tex]\(0 < 12\)[/tex] is true.
- So, the region that satisfies [tex]\(3x - 2y < 12\)[/tex] is the region below the line [tex]\(y = \frac{3}{2}x - 6\)[/tex].
### Finding the Intersection of the Regions
We now seek the overlapping region that satisfies both inequalities:
1. Graphical Solution:
- Draw the lines [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex] and [tex]\(y = \frac{3}{2}x - 6\)[/tex] on the same graph.
- Identify the region that is both above the line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex] and below the line [tex]\(y = \frac{3}{2}x - 6\)[/tex].
2. Points of Intersection:
- Find where the lines intersect by solving the system:
[tex]\[ -\frac{1}{4}x + 2.5 = \frac{3}{2}x - 6 \][/tex]
Multiply through by 4 to clear fractions:
[tex]\[ -x + 10 = 6x - 24 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 10 + 24 = 7x \quad \Rightarrow \quad 34 = 7x \quad \Rightarrow \quad x = \frac{34}{7} \approx 4.857 \][/tex]
Substitute [tex]\(x = \frac{34}{7}\)[/tex] back into either line equation to find [tex]\(y\)[/tex]:
[tex]\[ y = -\frac{1}{4}\left(\frac{34}{7}\right) + 2.5 = \frac{-34}{28} + 2.5 = -\frac{17}{14} + 2.5 = -\frac{17}{14} + \frac{35}{14} = \frac{18}{14} = \frac{9}{7} \approx 1.286 \][/tex]
Therefore, the lines intersect at approximately [tex]\(\left(\frac{34}{7}, \frac{9}{7}\right)\)[/tex].
### Conclusion
The solution to the system of inequalities is the region in the [tex]\(xy\)[/tex]-plane that:
- Lies above the line [tex]\(y = -\frac{1}{4}x + 2.5\)[/tex]
- Lies below the line [tex]\(y = \frac{3}{2}x - 6\)[/tex]
Graphing these regions and finding their overlap will visually provide the area satisfying both conditions.
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