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Simple Equations (Vedic Mathematics)

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Vedic Mathematics provides various concise methods to solve different types of equations − both linear and quadratic! These methods are extremely useful, especially in those cases, where the conventional methods take a lot of time.
These methods are derived from the “Sunyam Samyasamuccaye Sutra” of Vedic Mathematics.
Let us first consider the equation (x + a) (x + b) = (x + c) (x + d), where ab = cd, i.e., where the product of the constant terms on the Left Hand Side equals that on the Right Hand Side. Watch the following video to understand how to solve such an equation.
Solved Examples
Example 1:
Solve (x + 14) (x + 2) = (x + 28) (x + 1)
Solution:
Here, 14 × 2 = 28 × 1 = 28
⇒ x = 0
Example 2:
Solve (x + 21) (x + 3) = (x + 9) (x + 7)
Solution:
Here, 21 × 3 = 9 × 7 = 63
⇒ x = 0
Summary:
If, in the equation (x + a) (x+ b) = (x + c) (x + d), ab = cd, then the root of the equation will be x = 0.
Let us now consider the equation and watch the following video to understand the method to solve such equations.

Solved Examples

Example 1:

Solve

Solution:

(x + 4) + (x − 5) = 0

⇒ 2x − 1 = 0

⇒ x =

Example 2:

Solve

Solution:

(2x + 16) + (3x − 42) = 0

⇒ 5x − 26 = 0

Summary:

⇒ (ax + b) + (cx + d) = 0

Let us now consider the equation , where (ax + b) + (px + q) = (cx + d) + (rx + s), i.e., the sum of the numerators equals that of the denominators. Watch the following video to understand how to solve such an equation:

Solved Examples
Example 1:
Solve
Solution:
N1 = (4x − 8); N2 = (5x − 1)
N1 + N2 = (4x − 8) + (5x − 1) = 9x − 9
D1 = (2x + 3); D2 = (7x − 12)
D1 + D2 = (2x + 3) + (7x − 12) = 9x − 9
∴N1 + N2 = D1 + D2 = 9x − 9
⇒ 9x − 9 = 0
⇒ x = 1
N1 − D1 = 0
⇒ (4x − 8) − (2x + 3) = 0
⇒ 2x − 11 = 0
⇒
Example 2:
Solve
Solution:
N1 = (3x + 2); N2 = (2x − 3)
N1 + N2 = (3x + 2) + (2x − 3) = 5x − 1
D1 = (2x + 4); D2 = (3x − 5)
D1 + D2 = (2x + 4) + (3x − 5) = 5x − 1
∴N1 + N2 = D1 + D2 = 5x − 1
⇒ 5x − 1 = 0

N1 − D1 = 0
⇒ (3x + 2) − (2x + 4) = 0
⇒ x − 2 = 0
⇒ x = 2
Summary:
For the equation , where the sum of the numerators equals the sum of the denominators, i.e., where (ax + b) + (px + q) = (cx + d) + (rx + s), the roots of the equation are given as:
Root I: (ax + b) + (px + q) = 0

Root II: (ax + b) − (cx + d) = 0

Lastly, we come to the equation , where the sum of the denominators on the Left Hand Side equals that on the Right Hand Side, i.e., where(x + a) + (x + b) = (x + c) + (x + d). The following video will help you understand how to solve such equations:
Solved Examples
Example 1:
Solve
Solution:
D1 + D2 =
D3 + D4 =
∴D1 + D2 = D3 + D4

Example 2:
Solve
Solution:

Summary:
If and (x + a) + (x + b) = (x + c) + (x + d), then the root of the equation will be
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