Question

1. $y < 2x + 4$

$y \geq x + 1$

Explanation:

Step1: Graph \( y = 2x + 4 \)

The line \( y = 2x + 4 \) has a slope of \( 2 \) and a y - intercept of \( 4 \). Since the inequality is \( y<2x + 4 \), we draw a dashed line (because the inequality is strict, \( y
eq2x + 4 \)) and shade the region below the line.

Step2: Graph \( y=x + 1 \)

The line \( y=x + 1 \) has a slope of \( 1 \) and a y - intercept of \( 1 \). Since the inequality is \( y\geq x + 1 \), we draw a solid line (because the inequality is non - strict, \( y = x+1 \) is included in the solution set) and shade the region above the line.

Step3: Find the Intersection Region

The solution to the system of inequalities is the region that is shaded both below \( y = 2x+4 \) (dashed line) and above \( y=x + 1 \) (solid line). To find the intersection point of the two lines \( y = 2x+4 \) and \( y=x + 1 \), we set \( 2x+4=x + 1 \). Solving for \( x \): \( 2x-x=1 - 4\), so \( x=- 3 \). Substituting \( x = - 3 \) into \( y=x + 1 \), we get \( y=-3 + 1=-2 \). So the two lines intersect at \( (-3,-2) \). The solution region is bounded by the two lines, with the dashed line \( y = 2x + 4 \) (above the intersection point, the line \( y = 2x+4 \) is above \( y=x + 1 \)) and the solid line \( y=x + 1 \), and the region between them (below \( y = 2x+4 \) and above \( y=x + 1 \)).

Answer:

The solution is the region below the dashed line \( y = 2x+4 \) and above the solid line \( y=x + 1 \), with the intersection point at \((-3,-2)\). To graph it, draw a dashed line for \( y = 2x+4 \), shade below it, draw a solid line for \( y=x + 1 \), shade above it, and the overlapping shaded region is the solution.