Chapter 5 Review

Review Exercise 5 Solutions

Complete solved review worksheets covering MCQs, interval equations, system region boundaries, and bounded/unbounded linear programming solutions.

Question 1 Multiple Choice Questions
Choose the correct option. Click options to test your knowledge:
1. Which of the following is a linear equation?
a) 5x > 7
b) 4x - 2 < 1
c) 2x + 1 = 1
d) 4 = 1 + 3
2. The solution of 5x - 10 = 10 is:
a) 0
b) 50
c) 4
d) -4
3. If 7x + 4 < 6x + 6, then x belongs to the interval:
a) (2, ∞)
b) [2, ∞)
c) (-∞, 2)
d) (-∞, 2]
4. The linear equation formed out of the linear inequality is called:
a) Linear equation
b) Associated equation
c) Quadratic equation
d) Boundary equation
5. A statement involving symbols like <, >, ≤, ≥ is called:
a) Equation
b) Identity
c) Inequality
d) Formula
6. The degree of a linear inequality is:
a) 1
b) 2
c) 3
d) 0
7. The solution region restricted to the first quadrant is called:
a) Objective region
b) Feasible region
c) Corner region
d) Optimal region
8. A function that is to be maximized or minimized in linear programming is:
a) Solution function
b) Objective function
c) Feasible function
d) Corner function
Question 2 Solve Equations & Inequalities
Solve and represent the solution sets on a real number line:

(i) x + 52 = 1 - x

Solution
Multiply both sides by 2:
x + 5 = 2(1 - x)
x + 5 = 2 - 2x
Collect x terms:
x + 2x = 2 - 5
3x = -3 \implies x = -1
Solution Set: {-1}
Number Line (Solid circle at -1):
-3 -2 -1 0 1

(ii) 2x + 12 + 13 = 1 - x - 13

Solution
Multiply by LCM of denominators (which is 6):
3(2x + 1) + 2(1) = 6(1) - 2(x - 1)
6x + 3 + 2 = 6 - 2x + 2
6x + 5 = 8 - 2x
Solve:
6x + 2x = 8 - 5
8x = 3 \implies x = 3/8
Solution Set: {3/8} (which is 0.375).

(iii) 3x + 7 < 16

Solution
Subtract 7:
3x < 16 - 7 = 9
x < 3
Interval: (-∞, 3)

(iv) 5(x - 3) ≥ 26x - (10x + 4)

Solution
Expand both sides:
5x - 15 ≥ 26x - 10x - 4
5x - 15 ≥ 16x - 4
Solve:
-15 + 4 ≥ 16x - 5x
-11 ≥ 11x \implies x ≤ -1
Interval: (-∞, -1]
Number Line (Solid circle at -1, shading left):
-3 -2 -1 0 1
Question 3 Solution Regions (2D)
Find the solution region of the following linear inequality systems:

(i) 3x - 4y ≤ 12 and 3x + 2y ≥ 3

Solution
Boundary line 1: 3x - 4y = 12 (intercepts (4, 0), (0, -3)). Shaded side is origin-side (above/left).
Boundary line 2: 3x + 2y = 3 (intercepts (1, 0), (0, 3/2)). Shaded side is non-origin-side (above/right).

(ii) 2x + y ≤ 4 and x + 2y ≤ 6

Solution
Boundary line 1: 2x + y = 4 (intercepts (2, 0), (0, 4)). Shaded side is origin-side (below).
Boundary line 2: x + 2y = 6 (intercepts (6, 0), (0, 3)). Shaded side is origin-side (below).
Question 4 LP Maximization
Find the maximum value of g(x, y) = x + 4y subject to constraints:
x + y ≤ 4; x ≥ 0, y ≥ 0
Solution
Step 1: Feasible region:
Bounded in the first quadrant by line x + y = 4 (intercepts (4,0) and (0,4)).
Step 2: Corner points:
The corner points are O(0, 0), A(4, 0), and B(0, 4).
Step 3: Evaluate objective function:
Corner Point (x, y) g(x, y) = x + 4y Value
O(0, 0) 0 + 4(0) 0
A(4, 0) 4 + 4(0) 4
B(0, 4) [Optimal Max] 0 + 4(4) 16
The maximum value is 16 at corner point B(0, 4).
Question 5 LP Minimization
Find the minimum value of f(x, y) = 3x + 5y subject to constraints:
x + 3y ≥ 3; x + y ≥ 2; x ≥ 0, y ≥ 0
Solution
Step 1: Graph constraints:
• Line 1: x + 3y = 3 (intercepts (3, 0), (0, 1)).
• Line 2: x + y = 2 (intercepts (2, 0), (0, 2)).
Shading for both is away from the origin (above/right).
Step 2: Find intersection E:
Subtract equations:
(x + 3y) - (x + y) = 3 - 2
2y = 1 \implies y = 0.5
Substitute y = 0.5 \implies x = 2 - 0.5 = 1.5.
Intersection point is E(1.5, 0.5).
Step 3: Evaluate corner points:
Corner Point (x, y) f(x, y) = 3x + 5y Value
A(3, 0) 3(3) + 5(0) 9
D(0, 2) 3(0) + 5(2) 10
E(1.5, 0.5) [Optimal Min] 3(1.5) + 5(0.5) = 4.5 + 2.5 7
The minimum value is 7 at corner point E(1.5, 0.5).
Interactive Sandbox Vocabulary Quiz & Level Curve Solver
Test your vocabulary knowledge or slide the value C of the objective level curve to see it sweep across the feasible region and maximize visually!
Active Question