\( \leftarrow \quad \begin{array}{l}\text { Graph the feasible region for the } \\ \text { following system of inequalities. } \\ \begin{array}{l}x+y \leq 2 \\ x-y \geq 6\end{array}\end{array} \)

We need to graph the set of all points (x, y) that satisfy   (1) x + y ≤ 2   (2) x – y ≥ 6 Step 1. Rewrite the inequalities as statements about y. • For inequality (1), solve for y:   x + y ≤ 2 ⟹ y ≤ 2 – x.   The boundary line is y = 2 – x. • For inequality (2), solve for y:   x – y ≥ 6 ⟹ –y ≥ 6 – x ⟹ y ≤ x – 6.   The boundary line is y = x – 6. Step 2. Graph the boundary lines. • Graph y = 2 – x.   This line has an x-intercept at (2, 0) (set y = 0: 0 = 2 – x ⟹ x = 2) and a y-intercept at (0, 2). • Graph y = x – 6.   This line has an x-intercept at (6, 0) (set y = 0: 0 = x – 6 ⟹ x = 6) and a y-intercept at (0, –6). Since both inequalities have “≤”, the lines themselves are part of the solution. Step 3. Determine which side of each line to shade. • For y = 2 – x, the inequality is y ≤ 2 – x.   Pick a test point not on the line (a common choice is (0, 0)):    0 ≤ 2 – 0 ⟹ 0 ≤ 2, which is true.   So, shade the region below (or on) the line y = 2 – x. • For y = x – 6, the inequality is y ≤ x – 6.   Again, use (0, 0):    0 ≤ 0 – 6 ⟹ 0 ≤ -6, which is false.   So, the region that does NOT contain (0,0) is the solution. That is, we shade the region below (or on) the line y = x – 6. Step 4. Find the intersection (corner point) of the two boundary lines. Set the right sides equal:   2 – x = x – 6   Add x to both sides:  2 = 2x – 6   Add 6 to both sides:  8 = 2x   Divide by 2:  x = 4   Substitute x = 4 into y = 2 – x:  y = 2 – 4 = –2 So the lines intersect at (4, –2). Step 5. Describe the feasible region. The feasible region is the intersection of the two half-planes:   • All points on or below the line y = 2 – x   • And all points on or below the line y = x – 6 Thus, the solution set (feasible region) is all (x, y) such that   y ≤ min{2 – x, x – 6}. This region is unbounded. For x < 4 the inequality y ≤ x – 6 is more restrictive, and for x > 4 the inequality y ≤ 2 – x is more restrictive. The “corner” where the two lines meet is at (4, –2). Step 6. Sketch the graph. 1. Draw a coordinate plane. 2. Plot the line y = 2 – x by marking its intercepts (2, 0) and (0, 2) and draw a solid line. 3. Plot the line y = x – 6 by marking its intercepts (6, 0) and (0, –6) and draw a solid line. 4. Shade the region below each line.   • For y = 2 – x, shade everything below this line.   • For y = x – 6, shade everything below this line (note that the test point (0,0) should not be included for this inequality). 5. The feasible region is the overlapping (common) part of these shaded regions. It is the region that lies below both lines and includes the boundary lines. That is the final graph of the feasible region. To summarize, one acceptable answer is:   The feasible region for the system is    {(x, y) ∈ ℝ²: y ≤ 2 – x and y ≤ x – 6}, which is the area below (and including) the lines y = 2 – x and y = x – 6, with the lines intersecting at (4, –2).