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3-Phase Servo AVR (AC Voltage Stabilizer) — Parts, Tests, Repair & Maintenance

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3-Phase Servo AVR (AC Voltage Stabilizer) — Troubleshooting, Repair & Maintenance By Edwin Ogie • December 18, 2025 • -- AC Voltage Stabilizer — 3-phase servo control type (example from user photo) A practical, step-by-step guide to diagnose, repair and maintain 3-phase servo Automatic Voltage Regulators (AVR) / servo voltage stabilizers. Written in simple terms for technicians and maintenance teams working with generators, UPS rooms and factories. Includes videos, spare-parts list, safety checklist, troubleshooting flow and links to internal/external resources. Contents Why this matters In environments with unstable mains (frequent sags, surges or phase imbalance) a servo AVR protects sensitive equipment by continuously adjusting an autotransformer tap via a small servo motor. A well-maintained stabilizer saves equipment, reduces downtime and prevents costly damage. ...
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Simultaneous Equations part 1

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Lesson Note: Simultaneous Equations (Elimination, Substitution & Graphical Methods)

Simultaneous Equations: Methods of Solving

This lesson note explains how to solve a system of simultaneous equations using three methods: elimination, substitution, and graphical. Each method is broken down into simple, step‑by‑step worked examples.

1. Elimination Method

In the elimination method, you multiply (if needed) and add or subtract equations to cancel one variable.

  • Example 1:
    Equations: 2x + 3y = 13 and 3x - 2y = 4.
    1. Multiply the first equation by 2 and the second by 3 to obtain: → 4x + 6y = 26 and 9x - 6y = 12. 2. Add the equations: 4x + 9x = 13x and 26 + 12 = 38. 3. Solve: 13x = 38, so x = 38/13 ≈ 2.92. 4. Substitute x back into one equation to find y.
  • Example 2:
    Equations: 4x - y = 7 and 2x + y = 3.
    1. Add the two equations: (4x - y) + (2x + y) = 7 + 3, giving 6x = 10. 2. Solve: x = 10/6 = 5/3. 3. Substitute x back to solve for y.

2. Substitution Method

In the substitution method, solve one equation for one variable and then substitute that expression into the other equation.

  • Example 3:
    Equations: y = 2x - 1 and 3x + y = 9.
    1. Substitute y = 2x - 1 into 3x + y = 9. 2. This gives 3x + (2x - 1) = 9 → 5x - 1 = 9. 3. Solve: 5x = 10, so x = 2. 4. Then y = 2(2) - 1 = 3.
  • Example 4:
    Equations: y = x + 3 and 2x + 3y = 18.
    1. Substitute y = x + 3 into 2x + 3(x + 3) = 18. 2. Simplify: 2x + 3x + 9 = 18 → 5x + 9 = 18. 3. Solve: 5x = 9, so x = 9/5; then y = 9/5 + 3.

3. Graphical Method

In the graphical method, plot each equation on a coordinate plane. The point where the lines intersect is the solution to the system.

  • Example 5:
    Equations: y = 2x + 1 and y = -x + 4.
    1. Plot y = 2x + 1 (slope = 2, y–intercept = 1). 2. Plot y = -x + 4 (slope = -1, y–intercept = 4). 3. The intersection point of these lines gives the solution.
  • Example 6:
    Equations: y = x - 2 and y = -2x + 5.
    1. Plot y = x - 2 (slope = 1, y–intercept = -2). 2. Plot y = -2x + 5 (slope = -2, y–intercept = 5). 3. The intersection of these graphs is the solution.

Simultaneous Equations: Word Problem Examples

This section models real-life problems with simultaneous equations and shows how to solve them using the methods above.

Worked Examples

  • Example 7 (Elimination):
    Problem: A school sold 80 tickets for a play. Adult tickets cost ₦500 and child tickets cost ₦300. Total revenue was ₦34,000. Find the number of each ticket sold.
    1. Let x = adult tickets and y = child tickets. 2. Equation 1: x + y = 80. 3. Equation 2: 500x + 300y = 34,000. 4. Multiply Equation 1 by 300: 300x + 300y = 24,000. 5. Subtract: (500x - 300x) = 10,000, so 200x = 10,000, hence x = 50. 6. Then, y = 80 - 50 = 30.
  • Example 8 (Substitution):
    Problem: The sum of two numbers is 36, and the larger is 4 more than twice the smaller. Find the numbers.
    1. Let the smaller number = x and the larger = y. 2. Equation 1: x + y = 36. 3. Equation 2: y = 2x + 4. 4. Substitute Equation 2 into Equation 1: x + 2x + 4 = 36. 5. Solve: 3x = 32, so x ≈ 10.67; then y ≈ 25.33.
  • Example 9 (Graphical):
    Problem: A restaurant orders 50 fruits (apples and oranges). Apples cost ₦200 each and oranges ₦150 each. Total cost is ₦9,250. Find the quantities.
    1. Let x = apples and y = oranges. 2. Equation 1: x + y = 50. 3. Equation 2: 200x + 150y = 9,250. 4. Multiply Equation 1 by 150: 150x + 150y = 7,500. 5. Subtract: 50x = 1,750, so x = 35; then y = 50 - 35 = 15.
  • Example 10 (Elimination):
    Problem: Two numbers differ by 12 and sum to 60. Find them.
    1. Let the numbers be x and y (x > y). 2. Equation 1: x - y = 12. 3. Equation 2: x + y = 60. 4. Add the equations: 2x = 72, so x = 36; then y = 60 - 36 = 24.
  • Example 11 (Substitution):
    Problem: 3 pens and 2 pencils cost ₦1,200; 2 pens and 4 pencils cost ₦1,000. Find the cost of a pen.
    1. Let p = cost of a pen and q = cost of a pencil. 2. Equation 1: 3p + 2q = 1200. 3. Equation 2: 2p + 4q = 1000.
    4. Divide Equation 2 by 2: p + 2q = 500.
    5. Subtract from Equation 1: 2p = 700, so p = 350.

JAMB CBT Quiz on Simultaneous Equations

Total time: 900 seconds

This topic covers solving simultaneous equations using elimination, substitution, and graphical methods. Elimination Example: Multiply and add equations to cancel a variable. Substitution Example: Solve one equation for a variable and substitute into the other. Graphical Example: Plot both equations and identify the intersection. This topic demonstrates how to model real-life situations with simultaneous equations. Ticket sales problem using cost and quantity equations. Cost problem involving pens and pencils.

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