This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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To solve the system of inequalities, we need to find the region in the coordinate plane that satisfies all given conditions. This region is called the feasible region. The given system of inequalities is: 1) 2x + 3y 6 2) x - 2y -3 3) x 0 4) y 0 Step 1: Graph the boundary line for each inequality. For inequality 1: 2x + 3y 6 The boundary line is 2x + 3y = 6. To find two points on this line: • If x=0, then 3y=6 y=2. Point: (0,2). • If y=0, then 2x=6 x=3. Point: (3,0). Plot these points and draw a solid line connecting them. To determine the shaded region, test a point not on the line, for example, (0,0): 2(0) + 3(0) = 0 6. This is true, so the region satisfying 2x+3y 6 is below or on the line 2x+3y=6. For inequality 2: x - 2y -3 The boundary line is x - 2y = -3. To find two points on this line: • If x=0, then -2y=-3 y=(3)/(2). Point: (0, 1.5). • If y=0, then x=-3. Point: (-3,0). Plot these points and draw a solid line connecting them. To determine the shaded region, test a point not on the line, for example, (0,0): 0 - 2(0) = 0 -3. This is true, so the region satisfying x-2y -3 is above or on the line x-2y=-3. For inequality 3: x 0 This represents all points to the right of or on the y-axis. For inequality 4: y 0 This represents all points above or on the x-axis. Step 2: Identify the feasible region. The feasible region is the area where all four shaded regions overlap. Since x 0 and y 0, the feasible region will be restricted to the first quadrant. Step 3: Find the vertices of the feasible region. The vertices are the intersection points of the boundary lines that define the feasible region. 1. Intersection of x=0 and y=0: (0,0) 2. Intersection of x=0 and x-2y=-3: Substitute x=0 into x-2y=-3: 0 - 2y = -3 y = (3)/(2) (0, (3)/(2)) 3. Intersection of y=0 and 2x+3y=6: Substitute y=0 into 2x+3y=6: 2x + 3(0) = 6 2x = 6 x = 3 (3,0) 4. Intersection of 2x+3y=6 and x-2y=-3: From the second equation, x = 2y-3. Substitute this into the first equation: 2(2y-3) + 3y = 6 4y - 6 + 3y = 6 7y - 6 = 6 7y = 12 y = (12)/(7) Now substitute y=(12)/(7) back into x=2y-3: x = 2((12)/(7)) - 3 = (24)/(7) - (21)/(7) = (3)/(7) ((3)/(7), (12)/(7)) The solution to the system of inequalities is the feasible region, which is a polygon defined by its vertices. The vertices of the feasible region are: (0,0) (0, (3)/(2)) ((3)/(7), (12)/(7)) (3,0) The feasible region is the quadrilateral bounded by these four points.