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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 (Elimination, Substitution & Graphical Methods)

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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. The methods are explained step‑by‑step with clear worked examples, including word problems.

1. Elimination Method

In the elimination method, you add or subtract the equations (often after multiplying by suitable factors) 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 so that the y–coefficients become 6 and -6.
    → (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 of the original equations to find y.
  • Example 2:
    Equations: 4x - y = 7 and 2x + y = 3.
    1. Add the two equations to eliminate y: (4x - y) + (2x + y) = 7 + 3.
    2. This gives 6x = 10, so 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 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 both equations on a coordinate plane and identify the intersection point, which is the solution.

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

Simultaneous Equations: Word Problem Examples

Word problems can be modeled using simultaneous equations. Here are several examples solved using elimination or substitution.

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, 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, the larger = y.
    2. Equation 1: x + y = 36.
    3. Equation 2: y = 2x + 4.
    4. Substitute: x + 2x + 4 = 36 → 3x = 32, so x ≈ 10.67.
    5. Then y ≈ 2(10.67) + 4 = 25.33.
  • Example 9 (Graphical):
    Problem: A restaurant orders 50 fruits (apples and oranges). Apples cost ₦200 each and oranges ₦150 each. The total cost is ₦9,250. Find the quantities.
    1. Let x = apples, y = oranges.
    2. Equation 1: x + y = 50.
    3. Equation 2: 200x + 150y = 9,250.
    4. Solve by elimination or graphing. Using elimination, 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, with 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 and a pencil.
    1. Let p = cost of a pen, q = cost of a pencil.
    2. Equation 1: 3p + 2q = 1200.
    3. Equation 2: 2p + 4q = 1000. Divide Equation 2 by 2: p + 2q = 500.
    4. Subtract: (3p - p) = 1200 - 500 → 2p = 700, so p = 350.
    5. Then, 350 + 2q = 500 gives 2q = 150, so q = 75.

JAMB CBT Quiz on Simultaneous Equations

Total time: 900 seconds

This topic covers solving simultaneous equations using elimination, substitution, and graphical methods. Elimination: Solve 2x+3y=13 and 3x-2y=4 by making the y–coefficients opposites. Substitution: Solve y=2x-1 and 3x+y=9 by substituting for y. Graphical: Plot y=2x+1 and y=-x+4 and find their intersection. This topic shows how word problems can be modeled using simultaneous equations. Ticket sales problem modeling with adult and child tickets. Cost problem with pens and pencils.

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