Questions: 2x + y < 5 2x + y ≥ -3 3x - y > -4 3x - y ≤ 3

Solution Steps

We are given the following system of inequalities to solve: \[ \begin{align*}

1. & \quad 2x + y < 5 \\
2. & \quad 2x + y \geq -3 \\
3. & \quad 3x - y > -4 \\
4. & \quad 3x - y \leq 3 \end{align*} \]

We will rearrange the inequalities into standard form \( Ax + By \leq C \):

1. \( 2x + y < 5 \) becomes \( 2x + y - 5 < 0 \)
2. \( 2x + y \geq -3 \) becomes \( 2x + y + 3 \geq 0 \)
3. \( 3x - y > -4 \) becomes \( 3x - y + 4 > 0 \)
4. \( 3x - y \leq 3 \) becomes \( 3x - y - 3 \leq 0 \)

To analyze the boundaries of the inequalities, we convert them into equations:

1. \( 2x + y = 5 \)
2. \( 2x + y = -3 \)
3. \( 3x - y = -4 \)
4. \( 3x - y = 3 \)

We will solve the equations pairwise to find the intersection points:

1. Solve \( 2x + y = 5 \) and \( 2x + y = -3 \):

   • These lines are parallel and do not intersect, indicating no solution for this pair.

2. Solve \( 3x - y = -4 \) and \( 3x - y = 3 \):

   • These lines are also parallel and do not intersect, indicating no solution for this pair.

3. Solve \( 2x + y = 5 \) and \( 3x - y = -4 \):

   • Adding these equations gives \( 5x = 1 \) or \( x = \frac{1}{5} \).
   • Substituting \( x = \frac{1}{5} \) into \( 2x + y = 5 \) gives \( y = 5 - 2 \cdot \frac{1}{5} = \frac{23}{5} \).
   • Thus, one intersection point is \( \left( \frac{1}{5}, \frac{23}{5} \right) \).

4. Solve \( 2x + y = 5 \) and \( 3x - y = 3 \):

   • Adding these equations gives \( 5x = 8 \) or \( x = \frac{8}{5} \).
   • Substituting \( x = \frac{8}{5} \) into \( 2x + y = 5 \) gives \( y = 5 - 2 \cdot \frac{8}{5} = -\frac{6}{5} \).
   • Thus, another intersection point is \( \left( \frac{8}{5}, -\frac{6}{5} \right) \).

The feasible region is determined by the inequalities. We need to check which of the intersection points satisfy all the inequalities:

• Check \( \left( \frac{1}{5}, \frac{23}{5} \right) \):

  • \( 2 \cdot \frac{1}{5} + \frac{23}{5} < 5 \) (True)
  • \( 2 \cdot \frac{1}{5} + \frac{23}{5} \geq -3 \) (True)
  • \( 3 \cdot \frac{1}{5} - \frac{23}{5} > -4 \) (False)
  • \( 3 \cdot \frac{1}{5} - \frac{23}{5} \leq 3 \) (True)

• Check \( \left( \frac{8}{5}, -\frac{6}{5} \right) \):

  • \( 2 \cdot \frac{8}{5} - \frac{6}{5} < 5 \) (True)
  • \( 2 \cdot \frac{8}{5} - \frac{6}{5} \geq -3 \) (True)
  • \( 3 \cdot \frac{8}{5} + \frac{6}{5} > -4 \) (True)
  • \( 3 \cdot \frac{8}{5} + \frac{6}{5} \leq 3 \) (False)

Since neither intersection point satisfies all inequalities, we conclude that there is no solution to the system of inequalities.

Final Answer