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Find the solution region to the system of linear inequalities shown below. 3x + 4y < 12 x < 2 y > 1 -3 -2 -5-4 -3 -2 -1

Find the solution region to the system of linear inequalities shown below. 3x + 4y < 12 x < 2 y > 1 -3 -2 -5-4 -3 -2 -1

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Find the solution region to the system of linear inequalities shown below. (Shade the solution please)
**Transcription for Educational Website** **Task:** Find the solution region to the system of linear inequalities shown below. **Inequalities:** 1. \(3x + 4y < 12\) 2. \(x \leq 2\) 3. \(y \geq 1\) **Graph Description:** A coordinate plane is displayed with the x-axis ranging from -5 to 5 and the y-axis ranging from -5 to 5. The plane is divided into equal intervals with grid lines. The graph will be used to identify the solution region that satisfies all given inequalities. **Instructions:** 1. **Graph the inequality \(3x + 4y < 12\):** - Start by graphing the line \(3x + 4y = 12\). This is the boundary line of the region. - Since the inequality is \(<\), the region below this line will be shaded to indicate that those points are included. 2. **Graph the inequality \(x \leq 2\):** - Draw a vertical line at \(x = 2\). - Shade the area to the left of this line. 3. **Graph the inequality \(y \geq 1\):** - Draw a horizontal line at \(y = 1\). - Shade the area above this line. **Solution Region:** The solution region is the overlapping area that satisfies all three inequalities. This region should be clearly shaded on the graph. **Note:** Use dashed or solid lines appropriately based on the inequality symbols. For \(<\) or \(>\), use dashed lines, and for \(\leq\) or \(\geq\), use solid lines.
Transcribed Image Text:**Transcription for Educational Website** **Task:** Find the solution region to the system of linear inequalities shown below. **Inequalities:** 1. \(3x + 4y < 12\) 2. \(x \leq 2\) 3. \(y \geq 1\) **Graph Description:** A coordinate plane is displayed with the x-axis ranging from -5 to 5 and the y-axis ranging from -5 to 5. The plane is divided into equal intervals with grid lines. The graph will be used to identify the solution region that satisfies all given inequalities. **Instructions:** 1. **Graph the inequality \(3x + 4y < 12\):** - Start by graphing the line \(3x + 4y = 12\). This is the boundary line of the region. - Since the inequality is \(<\), the region below this line will be shaded to indicate that those points are included. 2. **Graph the inequality \(x \leq 2\):** - Draw a vertical line at \(x = 2\). - Shade the area to the left of this line. 3. **Graph the inequality \(y \geq 1\):** - Draw a horizontal line at \(y = 1\). - Shade the area above this line. **Solution Region:** The solution region is the overlapping area that satisfies all three inequalities. This region should be clearly shaded on the graph. **Note:** Use dashed or solid lines appropriately based on the inequality symbols. For \(<\) or \(>\), use dashed lines, and for \(\leq\) or \(\geq\), use solid lines.
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