Chapter 5: Linear Equations and Inequalities

Exercise 5.1 Solved Reference

High-fidelity, step-by-step solved reference guide for solving linear equations, linear inequalities, and graphing regions in the xy-plane, completely cleaned of all watermarks.

Question 1 Linear Equations
Solve each linear equation and represent its solution set on a real number line.

(i) 12x + 30 = -6

Solution
Write equation: 12x + 30 = -6
Isolate variable:
12x=-6 - 30 12x=-36 x=-36 / 12 = -3
Verification: Substitute x = -3:
12(-3) + 30 = -36 + 30 = -6 (True).
Answer: x = -3

(ii) x / 3 + 6 = -12

Solution
Isolate term:
x / 3=-12 - 6 x / 3=-18 x=-18 × 3 = -54
Answer: x = -54

(iii) x / 2 - 1 / 4 = 1 / 12

Solution
Add 1/4 to both sides:
x / 2=1/12 + 1/4 x / 2=(1 + 3) / 12 = 4 / 12 = 1 / 3 x=2 / 3
Answer: x = 2 / 3

(iv) 2 = 7(2x + 4) + 12x

Solution
Expand: 2 = 14x + 28 + 12x.
Simplify:
2=26x + 28 2 - 28=26x -26=26x \implies x = -1
Answer: x = -1

(v) (2x - 1) / 3 - 3x / 4 = 5 / 6

Solution
Multiply by LCM of denominators (12) to clear fractions:
4(2x - 1) - 3(3x)=2(5) 8x - 4 - 9x=10 -x - 4=10 -x=14 \implies x = -14
Answer: x = -14

(vi) -5x / 10 = 9 - 10x / 5

Solution
Simplify fractions: -x / 2 = 9 - 2x.
Isolate variable:
2x - x / 2=9 3x / 2=9 3x=18 \implies x = 6
Answer: x = 6
Question 2 Linear Inequalities
Solve each linear inequality and represent its solution set on a real number line.

(i) x - 6 < -2

Solution
Add 6 to both sides: x < 4.
The solution interval is (-∞, 4). Empty circle at 4.
Answer: x < 4

(ii) -9 > -16 + x

Solution
Add 16 to both sides: 7 > x \implies x < 7.
The solution interval is (-∞, 7). Empty circle at 7.
Answer: x < 7

(iii) 3 + 2x ≥ 3

Solution
Subtract 3 from both sides: 2x ≥ 0 \implies x ≥ 0.
The solution interval is [0, ∞). Filled circle at 0.
Answer: x ≥ 0

(iv) 6(x + 10) ≤ 0

Solution
Divide by 6: x + 10 ≤ 0 \implies x ≤ -10.
The solution interval is (-∞, -10]. Filled circle at -10.
Answer: x ≤ -10

(v) (8x - 3) / 3 - (9x - 4) / 4 < 2 / 12

Solution
Multiply both sides by LCM (12):
4(8x - 3) - 3(9x - 4)<2 32x - 12 - 27x + 12<2 5x<2 \implies x < 2 / 5
The solution set is x < 0.4. Empty circle at 2/5.
Answer: x < 2 / 5

(vi) (x - 2) / 4 ≤ -1 + x / 2

Solution
Multiply both sides by 4:
x - 2≤-4 + 2x 2≤x \implies x ≥ 2
The solution interval is [2, ∞). Filled circle at 2.
Answer: x ≥ 2
Question 3 Shading Regions in xy-Plane
Shade the solution region for the following linear inequalities in the xy-plane.

(i) 2x + y < 6

Solution
Associated boundary line equation: 2x + y = 6.
Find intercepts:
  • x-intercept (put y = 0): 2x = 6 \implies x = 3. Point: (3, 0).
  • y-intercept (put x = 0): y = 6. Point: (0, 6).
Test point (0, 0): 2(0) + 0 < 6 \implies 0 < 6 (True).
Shade the half-plane containing the origin. The boundary line is dashed (since inequality is strict).

(ii) 3x + 7y > 21

Solution
Associated boundary line equation: 3x + 7y = 21.
Find intercepts:
  • x-intercept: 3x = 21 \implies x = 7. Point: (7, 0).
  • y-intercept: 7y = 21 \implies y = 3. Point: (0, 3).
Test point (0, 0): 3(0) + 7(0) > 21 \implies 0 > 21 (False).
Shade the half-plane opposite to the origin. Boundary line is dashed.

(iii) 3x - 2y ≥ 6

Solution
Boundary line: 3x - 2y = 6.
Intercepts: (2, 0) and (0, -3).
Test point (0, 0): 0 ≥ 6 (False).
Shade the region opposite to origin. Boundary line is solid (includes boundary).

(iv) 5x - 4y ≤ 20

Solution
Boundary line: 5x - 4y = 20.
Intercepts: (4, 0) and (0, -5).
Test point (0, 0): 0 ≤ 20 (True).
Shade the upper half-plane containing origin. Line is solid.

(v) 2x + 1 > 0

Solution
Isolate x: x > -1/2.
Boundary line: vertical line x = -0.5.
Shade the region to the right of x = -0.5. Line is dashed.

(vi) 3y - 4 < 0

Solution
Isolate y: y < 4/3.
Boundary line: horizontal line y = 1.33.
Shade the region below y = 1.33. Line is dashed.
Question 4 Systems of Linear Inequalities
Indicate the solution region of the following linear inequality systems by shading.

(i) 2x - 3y ≤ 6 and 2x + 3y ≤ 12

Solution
For line 1: 2x - 3y = 6 has intercepts A(3, 0) and B(0, -2). Shading towards origin.
For line 2: 2x + 3y = 12 has intercepts C(6, 0) and D(0, 4). Shading towards origin.
The solution region is the intersection of these two shaded half-planes.

(ii) x + y ≥ 5 and -y + x ≤ 1

Solution
For line 1: x + y = 5 has intercepts (5, 0) and (0, 5). Shading away from origin.
For line 2: x - y = 1 has intercepts (1, 0) and (0, -1). Shading towards origin.
The solution region is the intersection of these two regions.

(iii) 3x + 7y ≥ 21 and x - y ≤ 2

Solution
For line 1: 3x + 7y = 21 has intercepts (7, 0) and (0, 3). Shading away from origin.
For line 2: x - y = 2 has intercepts (2, 0) and (0, -2). Shading towards origin.
The solution region is the intersection of these two regions.
Interactive Sandbox Linear Inequality & Number Line Grapher
Input parameters for the linear inequality ax + b < c to solve it algebraically and dynamically graph the solution region on the real number line below.
Algebraic Steps
Number Line Graph