Grapht \( \left\{\begin{array}{l}y<4 x+1 \\ y>-x+2\end{array}\right. \)

We wish to graph the system   y < 4x + 1   y > –x + 2 follow these steps: 1. Find the boundary lines:   • Line 1: y = 4x + 1   • Line 2: y = –x + 2   Note: Since the inequalities are strict (< and >), these lines themselves are not part of the solution region. When you draw them, use dashed lines. 2. Find the intersection point of the two lines:   Set 4x + 1 = –x + 2    4x + x = 2 – 1    5x = 1    x = 1/5   Then substitute back to find y:    y = 4(1/5) + 1 = 4/5 + 1 = 9/5   The lines intersect at (1/5, 9/5). 3. Decide which side of each line to shade.   • For y < 4x + 1:    Pick a test point not on the line (0,0) is convenient.    Test: 0 < 4(0) + 1 ⟹ 0 < 1, which is true.    Thus, the region containing (0,0) is the solution for y < 4x + 1.   • For y > –x + 2:    Again, test (0,0).    Test: 0 > –(0) + 2 ⟹ 0 > 2, which is false.    So pick another test point; try (0,3).    Test: 3 > –(0) + 2 ⟹ 3 > 2, which is true.    Thus, the region containing (0,3) is the solution for y > –x + 2. 4. Shade the common (overlapping) region:   • Shade below the dashed line y = 4x + 1   • Shade above the dashed line y = –x + 2   The solution to the system is the area where these two shaded regions overlap. 5. Sketching:   • Draw the two dashed boundary lines.   • Mark their intersection at (1/5, 9/5).   • Shade the region that lies below y = 4x + 1 and above y = –x + 2. This overlapping area is your final answer. This graph represents all points (x, y) that satisfy y < 4x + 1 and y > –x + 2.