Lessons 61–78: algebra foundations, factorisation, equations, and related reference methods in the new Vedic Math course.

Lesson 68 — General Two-Variable Equations by Cross-Multiplication

Technique explanation

The Urdhva-Tiryagbhyam sutra ("Vertically and crosswise") can be applied to solve any system of two linear equations of the form a₁x + b₁y = c₁ and a₂x + b₂y = c₂. This method uses a deterministic cross-product formula to find x and y directly.

The formula for x and y is derived from the cross-multiplication of coefficients and constants:

  • x = (c₁b₂ - c₂b₁) / (a₁b₂ - a₂b₁)
  • y = (a₁c₂ - a₂c₁) / (a₁b₂ - a₂b₁)

Method Condition: Reliable for any non-parallel linear system (where the denominator is non-zero).

Standard fallback: Cramer's Rule or the method of elimination.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Positive Coefficients

Solve: 3x + 4y = 10 and 5x + 2y = 12

  • Step 1: Calculate denominator (D): (3 2) - (5 4) = 6 - 20 = -14
  • Step 2: Calculate x: (10 2 - 12 4) / -14 = (20 - 48) / -14 = -28 / -14 = 2
  • Step 3: Calculate y: (3 12 - 5 10) / -14 = (36 - 50) / -14 = -14 / -14 = 1

Check: Verify with the second equation: 5(2) + 2(1) = 10 + 2 = 12. The result matches the constant term.

Example 2: Mixed Signs

Solve: 2x - 3y = 1 and 3x + 2y = 8

  • Step 1: Denominator (D): (2 2) - (3 -3) = 4 + 9 = 13
  • Step 2: x: (1 2 - 8 -3) / 13 = (2 + 24) / 13 = 26 / 13 = 2
  • Step 3: y: (2 8 - 3 1) / 13 = (16 - 3) / 13 = 13 / 13 = 1

Check: Verify with the second equation: 3(2) + 2(1) = 6 + 2 = 8. The result matches the constant term.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 67 — Solve by Addition and Subtraction

Technique explanation

The Sankalana-Vyavakalanabhyam sutra ("By addition and by subtraction") provides a rapid method for solving simultaneous linear equations where the coefficients of x and y are swapped between the two equations (e.g., ax + by = m and bx + ay = n).

By adding the two equations, we obtain a simplified equation for x + y. By subtracting them, we obtain a simplified equation for x - y. These two linear equations can then be solved easily to find the values of x and y.

Method Condition: Applies specifically when coefficients are swapped (a₁ = b₂ and b₁ = a₂).

Standard fallback: The method of elimination by equating coefficients or the substitution method.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Basic Swapped Coefficients

Solve for x and y: 31x + 23y = 85 and 23x + 31y = 77

  • Step 1: Add the equations: (31+23)x + (23+31)y = 85+77 → 54x + 54y = 162 → x + y = 3
  • Step 2: Subtract the equations: (31-23)x + (23-31)y = 85-77 → 8x - 8y = 8 → x - y = 1
  • Step 3: Solve x + y = 3 and x - y = 1: 2x = 4 → x = 2; 2y = 2 → y = 1

Check: Substitute x=2, y=1 into the second equation: 23(2) + 31(1) = 46 + 31 = 77. The solution is consistent.

Example 2: Large Coefficients

Solve for x and y: 103x + 97y = 303 and 97x + 103y = 297

  • Step 1: Add: 200x + 200y = 600 → x + y = 3
  • Step 2: Subtract: 6x - 6y = 6 → x - y = 1
  • Step 3: Solve: 2x = 4 → x = 2; 2y = 2 → y = 1

Check: Substitute x=2, y=1 into the second equation: 97(2) + 103(1) = 194 + 103 = 297. The solution is consistent.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 66 — Ratio Equations: When One Is in Ratio, the Other Is Zero

Technique explanation

The sutra Anurupye Shunyamanyat ("If one is in ratio, the other is zero") provides an instant solution for specific systems of simultaneous linear equations. If the ratio of the coefficients of one variable matches the ratio of the constant terms, then the other variable must be zero. This reduces the system to a single-variable equation that can be solved by simple division.

Standard fallback: The method of elimination or substitution to solve the system of simultaneous equations.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Solve 6x + 7y = 8 and 19x + 14y = 16
1. Check Ratios: y-coefficients are 7:14 (1:2). Constants are 8:16 (1:2).
2. Apply Sutra: Since the y-ratio matches the constant ratio, x = 0.
3. Solve for y: 7y = 8, so y = 8/7.
Result: x = 0, y = 8/7
Check: Substitute x = 0 and y = 8/7 into the second equation: 19(0) + 14(8/7) = 0 + 2 × 8 = 16. Match confirmed.

Example 2: Solve 12x + 78y = 12 and 16x + 96y = 16
1. Check Ratios: x-coefficients are 12:16 (3:4). Constants are 12:16 (3:4).
2. Apply Sutra: Since the x-ratio matches the constant ratio, y = 0.
3. Solve for x: 12x = 12, so x = 1.
Result: x = 1, y = 0
Check: Substitute x = 1 and y = 0 into the second equation: 16(1) + 96(0) = 16 + 0 = 16. Match confirmed.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 65 — Same-Sum Equations: When the Samuccaya Is Zero

Technique explanation

The sutra Shunyam Samyasamuccaye ("When the samuccaya is the same, that samuccaya is zero") provides an instant solution for specific rational equations. In equations of the form (x+a)/(x+b) = (x+c)/(x+d), if the sum of the numerator and denominator is the same on both sides (N1+D1 = N2+D2), that sum can be set to zero to find the root, avoiding complex cross-multiplication.

Standard fallback: Cross-multiplication, resulting in a quadratic equation (which often simplifies to linear if the x² terms cancel).

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Solve (x + 2)/(x + 3) = (x + 4)/(x + 1)
1. Check Samuccaya: (x + 2) + (x + 3) = 2x + 5. (x + 4) + (x + 1) = 2x + 5. They match.
2. Set to Zero: 2x + 5 = 0.
3. Solve: x = -2.5.
Result: x = -2.5
Check: Substitute x = -2.5 into the original equation: LHS = (-2.5 + 2)/(-2.5 + 3) = -0.5/0.5 = -1. RHS = (-2.5 + 4)/(-2.5 + 1) = 1.5/-1.5 = -1. Since -1 = -1, the solution is correct.

Example 2: Solve (x - 3)/(x + 5) = (x - 1)/(x + 3)
1. Check Samuccaya: (x - 3) + (x + 5) = 2x + 2. (x - 1) + (x + 3) = 2x + 2. They match.
2. Set to Zero: 2x + 2 = 0.
3. Solve: x = -1.
Result: x = -1
Check: Substitute x = -1 into the original equation: LHS = (-1 - 3)/(-1 + 5) = -4/4 = -1. RHS = (-1 - 1)/(-1 + 3) = -2/2 = -1. Since -1 = -1, the solution is correct.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 64 — Highest Common Factor of Polynomials

Technique explanation

The Highest Common Factor (HCF) of two polynomials must also divide their difference, according to the sutra Sankalana-Vyavakalanabhyam ("By addition and by subtraction"). By subtracting one polynomial from another to eliminate the highest degree term, we obtain a lower-degree expression that contains the HCF. Factoring this result and verifying the factors using the factor theorem identifies the HCF efficiently.

Standard fallback: The Euclidean Algorithm for polynomials, which involves repeated division and taking remainders until a remainder of zero is reached.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Find HCF of x³ + 6x² + 11x + 6 and x³ - x² - 10x - 8
1. Subtract: (x³ + 6x² + 11x + 6) - (x³ - x² - 10x - 8) = 7x² + 21x + 14.
2. Simplify: Divide by 7 to get x² + 3x + 2.
3. Factor: (x + 1)(x + 2). Both roots satisfy the originals.
Result: x² + 3x + 2
Check: Verify x² + 3x + 2 divides the first polynomial: (x² + 3x + 2)(x + 3) = x³ + 3x² + 2x + 3x² + 9x + 6 = x³ + 6x² + 11x + 6. Match confirmed.

Example 2: Find HCF of x³ + 7x² + 14x + 8 and x³ + 8x² + 19x + 12
1. Subtract: (x³ + 8x² + 19x + 12) - (x³ + 7x² + 14x + 8) = x² + 5x + 4.
2. Factor: (x + 1)(x + 4). Testing x = -1 and x = -4 confirms they are roots.
Result: x² + 5x + 4
Check: Verify x² + 5x + 4 divides the second polynomial: (x² + 5x + 4)(x + 3) = x³ + 5x² + 4x + 3x² + 15x + 12 = x³ + 8x² + 19x + 12. Match confirmed.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 63 — Argumental Division: Infer the Quotient Shape

Technique explanation

Argumental division applies the "first by first, last by last" principle to polynomial division. For exact division of a cubic by a linear factor, we can instantly determine the first and last terms of the quadratic quotient. The middle term is then found by comparing the coefficients of the dividend with the partial product of the known terms, bypassing the need for long division.

Standard fallback: Polynomial long division or synthetic division, where each power of x is reduced sequentially.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Divide 6x³ + 11x² - 4x - 4 by 2x + 1
1. First Term: 6x³ / 2x = 3x².
2. Last Term: -4 / 1 = -4.
3. Middle Term: In (2x + 1)(3x² + nx - 4), the x² term is 2nx² + 3x². Set 2n + 3 = 11, so n = 4.
Result: 3x² + 4x - 4
Check: Multiply (2x + 1)(3x² + 4x - 4) = 2x(3x² + 4x - 4) + 1(3x² + 4x - 4) = (6x³ + 8x² - 8x) + (3x² + 4x - 4) = 6x³ + 11x² - 4x - 4. Match confirmed.

Example 2: Divide x³ + 6x² + 11x + 6 by x + 1
1. First Term: x³ / x = x².
2. Last Term: 6 / 1 = 6.
3. Middle Term: In (x + 1)(x² + nx + 6), the x² term is nx² + x². Set n + 1 = 6, so n = 5.
Result: x² + 5x + 6
Check: Multiply (x + 1)(x² + 5x + 6) = x(x² + 5x + 6) + 1(x² + 5x + 6) = (x³ + 5x² + 6x) + (x² + 5x + 6) = x³ + 6x² + 11x + 6. Match confirmed.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 62 — Factorisation Checks at 1 and −1

Technique explanation

Vedic Mathematics uses the principles of Gunitasamuccayah and Gunakasamuccayah to verify factorisations. By evaluating a polynomial and its factors at x = 1 (sum of coefficients) and x = -1 (alternating sum), we can quickly detect errors. If the product of the sums of the factors does not equal the sum of the coefficients of the original polynomial, the factorisation is incorrect.

Standard fallback: Direct expansion of the factors to see if the resulting polynomial matches the original expression exactly.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Check x² + 7x + 12 = (x + 3)(x + 4)
1. At x = 1: Left side = 1 + 7 + 12 = 20. Right side = (1 + 3)(1 + 4) = 4 × 5 = 20. Check passes.
Result: Verified at x = 1.
Check: Expand (x + 3)(x + 4) = x(x + 4) + 3(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12. Match confirmed.

Example 2: Check x² - 5x + 6 = (x - 2)(x - 3)
1. At x = -1: Left side = (-1)² - 5(-1) + 6 = 1 + 5 + 6 = 12. Right side = (-1 - 2)(-1 - 3) = (-3) × (-4) = 12. Check passes.
Result: Verified at x = -1.
Check: Expand (x - 2)(x - 3) = x(x - 3) - 2(x - 3) = x² - 3x - 2x + 6 = x² - 5x + 6. Match confirmed.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 61 — Factor Quadratics: First by First, Last by Last

Technique explanation

The sutra Adyamadyenantyamantyena ("The first by the first and the last by the last") simplifies the factorisation of quadratics of the form ax² + bx + c. By identifying potential factor pairs for the leading coefficient a and the constant term c, we can determine the correct factors by ensuring their cross-products sum to the middle coefficient b. This method is most efficient when a and c have few factor pairs.

Standard fallback: The AC method (splitting the middle term). Multiply a and c, find factors that sum to b, and group the terms to factorise.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Factor 3x² + 8x + 4
1. Firsts: Factors of 3x² are 3x and x.
2. Lasts: Factors of 4 are (1, 4) or (2, 2).
3. Check Middle: (3x + 2)(x + 2) gives (3x × 2) + (x × 2) = 6x + 2x = 8x. Correct.
Result: (3x + 2)(x + 2)
Check: Expand (3x + 2)(x + 2) = 3x(x + 2) + 2(x + 2) = 3x² + 6x + 2x + 4 = 3x² + 8x + 4. Match confirmed.

Example 2: Factor 6x² - 5x - 6
1. Firsts: Factors of 6x² are (3x, 2x) or (6x, x).
2. Lasts: Factors of -6 include (2, -3).
3. Check Middle: (3x + 2)(2x - 3) gives (3x × -3) + (2x × 2) = -9x + 4x = -5x. Correct.
Result: (3x + 2)(2x - 3)
Check: Expand (3x + 2)(2x - 3) = 3x(2x - 3) + 2(2x - 3) = 6x² - 9x + 4x - 6 = 6x² - 5x - 6. Match confirmed.

Academy Course
Vedic Math: Foundations to Reference Applications
Vedic Math — Reference Algebra and Equations