41. \( \begin{array}{l}2 x-3 y>-6 \\ 5 x-3 y<3 \\ x+3 y>-3\end{array} \)

We wish to find all ordered pairs (x, y) that satisfy   (1) 2x – 3y > –6   (2) 5x – 3y < 3   (3) x + 3y > –3 One acceptable method is to “solve for y” in each inequality so that we can see the region in the xy‐plane. Step 1. Write each inequality in the form y < (or >) an expression in x. • For (1): 2x – 3y > –6   Subtract 2x: –3y > –6 – 2x   Multiply both sides by –1 (remember to reverse the inequality):    3y < 6 + 2x   Divide by 3:   y < (2/3)x + 2 • For (2): 5x – 3y < 3   Subtract 5x: –3y < 3 – 5x   Multiply by –1 (and reverse the inequality): 3y > 5x – 3   Divide by 3:   y > (5x – 3)/3 • For (3): x + 3y > –3   Subtract x:  3y > –3 – x   Divide by 3:   y > (–3 – x)/3  or y > –(x/3) – 1 So the inequalities become   y < (2/3)x + 2              [Upper bound]   y > (5x – 3)/3              [Lower bound from (2)]   y > –(x/3) – 1              [Lower bound from (3)] Step 2. Notice that we have two lower bounds for y. In order for a point (x, y) to satisfy both,   y must be greater than the larger of the two numbers  (5x – 3)/3  and –(x/3) – 1. That is, we must have   y > max { (5x – 3)/3, –(x/3) – 1 }    and  y < (2/3)x + 2. Step 3. It is useful now to “compare” the two lower bounds. Find x such that   (5x – 3)/3 = –(x/3) – 1. Multiply both sides by 3:   5x – 3 = –x – 3. Bring –x to the left:   5x + x – 3 = –3  ⟹ 6x – 3 = –3. Add 3: 6x = 0, so x = 0. Thus, for x = 0 the two lower bounds are equal. Now check a test value:  • If x > 0, say x = 1:   (5(1) – 3)/3 = 2/3 and –(1/3) – 1 = –4/3.   Since 2/3 > –4/3, for x > 0 the effective lower bound is (5x – 3)/3.  • If x < 0, say x = –1:   (5(–1) – 3)/3 = (–5 – 3)/3 = –8/3 ≈ –2.67,   –(–1/3) – 1 = 1/3 – 1 = –2/3 ≈ –0.67.   Now –0.67 > –2.67, so for x < 0 the effective lower bound is –(x/3) – 1. So we must treat the answer in two cases: Case 1. For x < 0:  The conditions become   • x < 0 (with x not too “small” – we will determine the x–range soon)   • Lower bound: y > –(x/3) – 1   • Upper bound: y < (2/3)x + 2 For these to be compatible we need   –(x/3) – 1 < (2/3)x + 2. Multiply everything by 3 to clear fractions:   –x – 3 < 2x + 6. Add x to both sides:   –3 < 3x + 6. Subtract 6:   –9 < 3x   ⟹ x > –3. Thus, in Case 1 we require –3 < x < 0. Case 2. For x > 0:  The conditions become   • x > 0,   • Lower bound: y > (5x – 3)/3,   • Upper bound: y < (2/3)x + 2. For these to be compatible we need   (5x – 3)/3 < (2/3)x + 2. Multiply by 3:   5x – 3 < 2x + 6. Subtract 2x:   3x – 3 < 6. Add 3:   3x < 9   ⟹ x < 3. Thus, in Case 2 we require 0 < x < 3. At x = 0 the two lower bounds coincide (both equal –1), so x = 0 fits into either description (with y > –1 and also y < 2, since (2/3)(0)+2 = 2). Step 4. Write the full answer. The solution set is all real numbers x and y satisfying  • For –3 < x < 0:  –(x/3) – 1 < y < (2/3)x + 2.  • For x = 0:     y satisfies –1 < y < 2.  • For 0 < x < 3:  (5x – 3)/3 < y < (2/3)x + 2. Any point (x, y) in the plane that satisfies these conditions will satisfy the original system. An acceptable final answer is:   {x, y ∈ ℝ such that –3 < x < 3 and    if x < 0, then –(x/3) – 1 < y < (2/3)x + 2, and    if x ≥ 0, then (5x – 3)/3 < y < (2/3)x + 2.} Note: The endpoints x = –3 and x = 3 are not included because when x = –3 or 3 the corresponding inequalities would force impossible conditions. Any answer that is algebraically equivalent to the one above is correct.