Answer :
To solve the system of inequalities:
[tex]\[ \left\{\begin{aligned} x + y & > 1 \\ 3x - 5 & \leq y \\ y & < 2x \end{aligned}\right. \][/tex]
Let's analyze and plot the region that satisfies all three inequalities step-by-step.
### Step 1: Graph each inequality individually
1. Inequality 1: \(x + y > 1\)
- The boundary line is \(x + y = 1\), which can be rearranged as \(y = -x + 1\). Plot the line \(y = -x + 1\).
- Since the inequality is \(>\), the solution region is above the line \(y = -x + 1\).
2. Inequality 2: \(3x - 5 \leq y\)
- The boundary line is \(y = 3x - 5\). Plot the line \(y = 3x - 5\).
- Since the inequality is \(\leq\), the solution region is on or above the line \(y = 3x - 5\).
3. Inequality 3: \(y < 2x\)
- The boundary line is \(y = 2x\). Plot the line \(y = 2x\).
- Since the inequality is \(<\), the solution region is below the line \(y = 2x\).
### Step 2: Find the intersections of the boundary lines
To find the region that satisfies all three inequalities, we need to identify the intersection points of the boundary lines and determine where all three conditions are met.
1. Intersection of \(y = -x + 1\) and \(y = 3x - 5\):
[tex]\[ -x + 1 = 3x - 5 \implies 1 + 5 = 3x + x \implies 6 = 4x \implies x = \frac{6}{4} = \frac{3}{2} \][/tex]
Substituting \(x = \frac{3}{2}\) into \(y = -x + 1\):
[tex]\[ y = -\frac{3}{2} + 1 = -\frac{3}{2} + \frac{2}{2} = -\frac{1}{2} \][/tex]
Intersection point: \(\left(\frac{3}{2}, -\frac{1}{2}\right)\)
2. Intersection of \(y = -x + 1\) and \(y = 2x\):
[tex]\[ -x + 1 = 2x \implies 1 = 3x \implies x = \frac{1}{3} \][/tex]
Substituting \(x = \frac{1}{3}\) into \(y = 2x\):
[tex]\[ y = 2\left(\frac{1}{3}\right) = \frac{2}{3} \][/tex]
Intersection point: \(\left(\frac{1}{3}, \frac{2}{3}\right)\)
3. Intersection of \(y = 3x - 5\) and \(y = 2x\):
[tex]\[ 3x - 5 = 2x \implies 3x - 2x = 5 \implies x = 5 \][/tex]
Substituting \(x = 5\) into \(y = 2x\):
[tex]\[ y = 2 \cdot 5 = 10 \][/tex]
Intersection point: \((5, 10)\)
### Step 3: Determine the feasible region
Plot the lines and mark the intersection points:
1. \(y = -x + 1\)
2. \(y = 3x - 5\)
3. \(y = 2x\)
Identify the region that satisfies all the inequalities:
1. Above \(y = -x + 1\)
2. On or above \(y = 3x - 5\)
3. Below \(y = 2x\)
### Step 4: Describe the feasible region
Upon plotting and analyzing, the feasible solution region is the area bounded by the lines and intersection points:
- The intersection points are \(\left(\frac{3}{2}, -\frac{1}{2}\right)\), \(\left(\frac{1}{3}, \frac{2}{3}\right)\), and \((5, 10)\).
- This triangular region satisfies all the inequalities.
### Summary
The solution to the system of inequalities is the region defined by:
[tex]\[ \left\{\begin{aligned} x + y & > 1 \\ 3x - 5 & \leq y \\ y & < 2x \end{aligned}\right. \][/tex]
This region can be visualized as the triangular area bounded by the intersection points described above.
[tex]\[ \left\{\begin{aligned} x + y & > 1 \\ 3x - 5 & \leq y \\ y & < 2x \end{aligned}\right. \][/tex]
Let's analyze and plot the region that satisfies all three inequalities step-by-step.
### Step 1: Graph each inequality individually
1. Inequality 1: \(x + y > 1\)
- The boundary line is \(x + y = 1\), which can be rearranged as \(y = -x + 1\). Plot the line \(y = -x + 1\).
- Since the inequality is \(>\), the solution region is above the line \(y = -x + 1\).
2. Inequality 2: \(3x - 5 \leq y\)
- The boundary line is \(y = 3x - 5\). Plot the line \(y = 3x - 5\).
- Since the inequality is \(\leq\), the solution region is on or above the line \(y = 3x - 5\).
3. Inequality 3: \(y < 2x\)
- The boundary line is \(y = 2x\). Plot the line \(y = 2x\).
- Since the inequality is \(<\), the solution region is below the line \(y = 2x\).
### Step 2: Find the intersections of the boundary lines
To find the region that satisfies all three inequalities, we need to identify the intersection points of the boundary lines and determine where all three conditions are met.
1. Intersection of \(y = -x + 1\) and \(y = 3x - 5\):
[tex]\[ -x + 1 = 3x - 5 \implies 1 + 5 = 3x + x \implies 6 = 4x \implies x = \frac{6}{4} = \frac{3}{2} \][/tex]
Substituting \(x = \frac{3}{2}\) into \(y = -x + 1\):
[tex]\[ y = -\frac{3}{2} + 1 = -\frac{3}{2} + \frac{2}{2} = -\frac{1}{2} \][/tex]
Intersection point: \(\left(\frac{3}{2}, -\frac{1}{2}\right)\)
2. Intersection of \(y = -x + 1\) and \(y = 2x\):
[tex]\[ -x + 1 = 2x \implies 1 = 3x \implies x = \frac{1}{3} \][/tex]
Substituting \(x = \frac{1}{3}\) into \(y = 2x\):
[tex]\[ y = 2\left(\frac{1}{3}\right) = \frac{2}{3} \][/tex]
Intersection point: \(\left(\frac{1}{3}, \frac{2}{3}\right)\)
3. Intersection of \(y = 3x - 5\) and \(y = 2x\):
[tex]\[ 3x - 5 = 2x \implies 3x - 2x = 5 \implies x = 5 \][/tex]
Substituting \(x = 5\) into \(y = 2x\):
[tex]\[ y = 2 \cdot 5 = 10 \][/tex]
Intersection point: \((5, 10)\)
### Step 3: Determine the feasible region
Plot the lines and mark the intersection points:
1. \(y = -x + 1\)
2. \(y = 3x - 5\)
3. \(y = 2x\)
Identify the region that satisfies all the inequalities:
1. Above \(y = -x + 1\)
2. On or above \(y = 3x - 5\)
3. Below \(y = 2x\)
### Step 4: Describe the feasible region
Upon plotting and analyzing, the feasible solution region is the area bounded by the lines and intersection points:
- The intersection points are \(\left(\frac{3}{2}, -\frac{1}{2}\right)\), \(\left(\frac{1}{3}, \frac{2}{3}\right)\), and \((5, 10)\).
- This triangular region satisfies all the inequalities.
### Summary
The solution to the system of inequalities is the region defined by:
[tex]\[ \left\{\begin{aligned} x + y & > 1 \\ 3x - 5 & \leq y \\ y & < 2x \end{aligned}\right. \][/tex]
This region can be visualized as the triangular area bounded by the intersection points described above.
