Saturday, July 25, 2020

Systems of Linear Inequalities - Practice Problem and Answer (#5)

Problem:

 Graph the system

 y ≤ 2x - 4
y > -x + 5

 Answer:
For y = 2x - 4: y-intercept: Put x = 0 into y = 2x - 4 to get y = 2(0) - 4 = -4. Thus, (0, -4) is a point on y = 2x - 4. x-intercept: Put y = 0 into y = 2x - 4 to get 0 = 2x - 4 implies 4 = 2x implies x = 2. Thus, (2,0) is a point on y = 2x - 4. For y = - x + 5: y-intercept: Put x = 0 into y = -x + 5 to get y = -0 + 5 = 5. Thus, (0, 5) is a point on y = -x+ 5. x-intercept: Put y = 0 into y = -x+5 to get 0 = -x+5 implies x = 5. Thus, (5, 0) is a point on y = -x+ 5. Where do y = 2x - 4 and y = -x + 5 intersect? Solve the system y = 2x - 4, y = -x+ 5. Thus, 2x - 4 = -x + 5 implies that 3x - 4 = 5 implies that 3x = 9 implies that x = 3. Put x = 3 into y = -x+5 to get y = -3 + 5 = 2. So, y = 2x - 4 and y = -x + 5 intersect at (3, 2). Now, put everything on the graph so far and label the regions I to IV.

Let region I be above y = -x+ 5 and to the right of y = 2x - 4. Let region II be above y = -x+ 5 and to the left of y = 2x - 4. Let region III be below y = -x + 5 and to the left of y = 2x - 4. Let region IV be below y = -x + 5 and to the right of y = 2x - 4. For region I: Try (6,0) and put it into the system at the start of the problem to get: 0 less than or equal to 2(6) - 4 = 8 and 0 > -6 + 5 = -1. Since both of these are true, we will shade region I. For region II, Try (0, 6) and put it into the system we started with. We get 6 less than or equal to 2(0) - 4 = -4 for one of the equations which is false. Thus, we will not shade region II. For region III, try (0,0) and put into the system of inequalities we started with to get 0 less than or equal to 2(0) - 4 = -4 for one of the equations in the system. Thus, we will not shade region III. For region IV, try (3,0) and put into the original system. By substitution into the system, we have that the bottom equation becomes 0 > -3 + 5 = 2 which is false.






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