W. Wraph the system \( y>3 x+1 \) and \( y<\frac{1}{2} x-4 \)

1. We are given the system of inequalities \[ y>3x+1 \] and \[ y<\frac{1}{2}x-4. \] 2. The inequality \( y > 3x+1 \) represents the region above the line \[ y=3x+1, \] while \( y < \frac{1}{2}x-4 \) represents the region below the line \[ y=\frac{1}{2}x-4. \] (Since both inequalities are strict, the boundary lines are not included in the solution.) 3. To have any point \((x,y)\) satisfying both inequalities simultaneously, the two boundaries must satisfy \[ 3x+1<\frac{1}{2}x-4. \] 4. Solve the inequality \( 3x+1<\frac{1}{2}x-4 \): - Multiply both sides by 2 to eliminate the fraction: \[ 2(3x+1) < 2\left(\frac{1}{2}x-4\right) \quad \Rightarrow \quad 6x+2 < x-8. \] - Subtract \( x \) from both sides: \[ 6x-x+2 < -8 \quad \Rightarrow \quad 5x+2 < -8. \] - Subtract 2 from both sides: \[ 5x < -10. \] - Divide both sides by 5: \[ x < -2. \] 5. For any \( x < -2 \), the corresponding \( y \) must satisfy \[ 3x+1 < y < \frac{1}{2}x-4. \] Note that when \( x=-2 \), both lines give \( y=-5 \) (since \( 3(-2)+1=-5 \) and \(\frac{1}{2}(-2)-4=-5\)); however, \( x=-2 \) is not in the solution set because the inequality \( 3x+1 < \frac{1}{2}x-4 \) must hold. 6. Therefore, the solution region of the system is \[ \{(x,y) \mid x < -2 \text{ and } 3x+1 < y < \frac{1}{2}x-4\}. \] 7. On a graph, you would: - Draw the line \( y=3x+1 \) as a dashed line and shade the region above it. - Draw the line \( y=\frac{1}{2}x-4 \) as a dashed line and shade the region below it. - The intersection of these two regions for \( x < -2 \) is the solution of the system.