Using the sutra Shunyam Samyaschahye, Solve the equation:1. (2x + 1) +

1.	Using the sutra Shunyam Samyaschahye, Solve the equation:1. (2x + 1) + (x + 3) = 5x + 42. a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)3. (x + 1)(x + 9) = (x + 3)(x + 3)4. x/2 + x/3 = x/4 + x/1.
Answer» 1. (2x + 1) + (x + 3) = 5x + 4 (2x + 1) + (x + 3) = 5x + 4 independent term of both sides are same (= 4) So, x = 0. 2. a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1) Here a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1) Thus, (x – 1) is common factor is both sides, then x – 1 = 0 ⇒ x = 1 3. (x + 1)(x + 9) = (x + 3)(x + 3) (x + 1)(x + 9) = (x + 3)(x + 3) Here independent term in both sides are same (= 9) then x = 0 4. x/2 + x/3 = x/4 + x/1. Numerator (x) is same in both sides Sum of denominators in R.H.S. = 2 + 3 = 5 Sum of denominators in R.H.S. = 4 + 1 = 5 Thus, According to formula, x = 0

Using the sutra Shunyam Samyaschahye, Solve the equation:1. (2x + 1) + (x + 3) = 5x + 42. a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)3. (x + 1)(x + 9) = (x + 3)(x + 3)4. x/2 + x/3 = x/4 + x/1.

Answer»

1. (2x + 1) + (x + 3) = 5x + 4

(2x + 1) + (x + 3) = 5x + 4

independent term of both sides are same (= 4)

So, x = 0.

2. a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)

Here

a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)

Thus, (x – 1) is common factor is both sides, then x – 1 = 0

⇒ x = 1

3. (x + 1)(x + 9) = (x + 3)(x + 3)

(x + 1)(x + 9) = (x + 3)(x + 3)

Here independent term in both sides are same (= 9) then x = 0

4. x/2 + x/3 = x/4 + x/1.

Numerator (x) is same in both sides

Sum of denominators in R.H.S. = 2 + 3 = 5

Sum of denominators in R.H.S. = 4 + 1 = 5

Thus, According to formula, x = 0