University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

30

Pages:

5
Author:

Herbosambx

Question 1 Graphically represent the solution set of a system of equations: 𝑥 + 𝑦 > 2 x+y>2 𝑥 + 𝑦 < − 3 x+y<−3 Answer 1 Graph of the solution of the inequality 𝑥 + 𝑦 > 2 x+y>2 is the region above the line 𝑥 + 𝑦 = 2 x+y=2. The points on the line 𝑥 + 𝑦 = 2 x+y=2 are not solutions of the inequality as it is a strict inequality. So, we draw a dotted line for the line 𝑥 + 𝑦 = 2 x+y=2. Similarly, the graph of the solution of the inequality 𝑥 + 𝑦 < − 3 x+y<−3 is the region below the line 𝑥 + 𝑦 = − 3 x+y=−3. The points on the line 𝑥 + 𝑦 = − 3 x+y=−3 are not solutions of the inequality as it is a strict inequality. So, we draw a dotted line for the line 𝑥 + 𝑦 = − 3 x+y=−3. Graph attached. Question 2 Find the slope and 𝑦 y-intercept of the line. Graph the line 5 𝑥 + 𝑦 = 5 5x+y=5. Answer 1 The slope-intercept form is 𝑦 = 𝑚 𝑥 + 𝑏 y=mx+b, where 𝑚 m is the slope and 𝑏 b is the 𝑦 y-intercept. Given the straight line 5 𝑥 + 𝑦 = 5 5x+y=5, it can be written as 𝑦 = − 5 𝑥 + 5 y=−5x+5. Now comparing the above equation with the slope-intercept form 𝑦 = 𝑚 𝑥 + 𝑐 y=mx+c, we get: slope = 𝑚 = − 5 and 𝑦 -intercept = 𝑐 = 5. slope=m=−5andy-intercept=c=5. Graph attached. Question 3 Graph the solution set of each system of inequalities: 𝑥 x is greater than or equal to 0 0. 𝑥 + 𝑦 x+y is less than or equal to 4 4. 2 𝑥 + 𝑦 2x+y is less than or equal to 5 5. Answer 1 For the inequalities, we make auxiliary equations: 𝑥 = 0 x=0 𝑥 + 𝑦 = 4 x+y=4 2 𝑥 + 𝑦 = 5 2x+y=5. Step 1: For 𝑥 ≥ 0 x≥0, shade the right side of 𝑥 = 0 x=0. For 𝑥 + 𝑦 ≤ 4 x+y≤4, shade toward ( 0 , 0 ) (0,0) because 𝑥 = 0 , 𝑦 = 0 x=0,y=0 satisfies the inequality. For 2 𝑥 + 𝑦 ≤ 5 2x+y≤5, shade toward ( 0 , 0 ) (0,0) because 𝑥 = 0 , 𝑦 = 0 x=0,y=0 satisfies the inequality. The common region is the shaded region in black. Graph and stepwise explanation attached. Question 4 Determine whether the ordered pair is a solution to the system of inequalities 𝑥 + 𝑦 < 4 x+y<4, 𝑦 ≤ 2 𝑥 + 1 y≤2x+1: a. ( 0 , 1 ) b. ( 3 , 1 ) c. ( 2 , 0 ) d. ( 1 , 4 ) . a. (0,1)b. (3,1)c. (2,0)d. (1,4). Answer 1 Step 1: Graph the inequalities. Given inequalities: 𝑥 + 𝑦 < 4 , 𝑦 ≤ 2 𝑥 + 1. x+y<4,y≤2x+1. Step 2: Check the boundary points. At ( 1 , 3 ) (1,3): 1 + 3 < 4 ⇒ not true . 1+3<4⇒not true. 3 ≤ 2 ( 1 ) + 1 ⇒ true . 3≤2(1)+1⇒true. Therefore, ( 1 , 3 ) (1,3) is not a solution. At ( 0 , 1 ) (0,1): 0 + 1 < 4 ⇒ true . 0+1<4⇒true. 1 ≤ 2 ( 0 ) + 1 ⇒ true . 1≤2(0)+1⇒true. Therefore, ( 0 , 1 ) (0,1) is a solution. Graph attached. Question 5 Graph the solution set of the system of inequalities or indicate that the system has no solution: 𝑥 + 𝑦 > 4 , 𝑥 + 𝑦 < − 1. x+y>4,x+y<−1. Answer 1 A system of inequalities will have solutions if they have a common region when graphed. From the given equations of inequalities, we can find that they contradict each other. To graph the inequalities, first convert them into standard equations: 𝑥 + 𝑦 = 4 , 𝑥 + 𝑦 = − 1. x+y=4,x+y=−1. Here, both the equations are parallel to each other. From the graph, we find that there is no region which is common to both. It means that the given system of inequalities has no solutions. Graph attached.