Solve each of the following equations: x 3 + x 4 = 42

Given equation is x 3 + x 4 = 42 . We have to find the value of x . We have, x 3 + x 4 = 42 L.C.M. of 3 and 4 is 12 . 4 x + 3 x 12 = 42 7 x 12 = 42 Multiplying both sides by 12 . 7 x 12 × 12 = 42 × 12 7 x = 504 Dividing both sides by 7 . 7 x 7 = 504 7 x = 72 Therefore, the required value is x = 72 .

Solve each of the following equations: 5 + 1 3 x = 2 x

The given equation is 5 + 1 3 x = 2 x . We have to find the value of x . We have, 5 + 1 3 x = 2 x Transposing + 1 3 x to R.H.S. 5 = 2 x - 1 3 x L.C.M. of 1 and 3 is 3 . 5 = 6 x - 2 x 3 5 = 5 x 3 Multiplying both sides by 3 . 5 × 3 = 5 x 3 × 3 15 = 5 x Dividing both sides by 5 . 15 5 = 5 x 5 ⇒ x = 3 Therefore, the required value is x = 3 .

Solve each of the following equations: x - 3 2 - 4 = 9

The given equation is x - 3 2 - 4 = 9 . We have to find the value of x . We have, x - 3 2 - 4 = 9 Transposing - 4 to R.H.S. x - 3 2 = 9 + 4 x - 3 2 = 13 Multiplying both sides by 2 . x - 3 2 × 2 = 13 × 2 x - 3 = 26 Transposing - 3 to R.H.S. x = 26 + 3 ⇒ x = 29 Therefore, the required value is x = 29 .

Solve each of the following equations: y - 6 2 = 3 - 2 y 5

The given equation is y - 6 2 = 3 - 2 y 5 . We have to find the value of y . We have, y - 6 2 = 3 - 2 y 5 Transposing 3 - 2 y 5 to L.H.S. y - 6 2 - 3 - 2 y 5 = 0 L.C.M. of 2 and 5 is 10 . 5 y - 6 - 2 3 - 2 y 10 = 0 Multiplying both sides by 10 . 5 y - 6 - 2 3 - 2 y 10 × 10 = 0 × 10 5 y - 30 - 6 + 4 y = 0 9 y - 36 = 0 We have 9 y - 36 = 0 . Transposing - 36 to R.H.S. 9 y = 0 + 36 Dividing both sides by 9 . 9 y 9 = 36 9 ⇒ y = 4 Therefore, the required value is y = 4 .

Solve each of the following equations: 2 x - 3 = 1 3 x + 1 3

The given equation is 2 x - 3 = 1 3 x + 1 3 . We have to find the value of x . We have, 2 x - 3 = 1 3 x + 1 3 Transposing - 3 to R.H.S. and + 1 3 x to L.H.S. 2 x - 1 3 x = 1 3 + 3 L.C.M. of 1 and 3 is 3 . 6 x - x 3 = 1 + 9 3 5 x 3 = 10 3 Multiplying both sides by 3 . 5 x 3 × 3 = 10 3 × 3 5 x = 10 Dividing both sides by 5 . 5 x 5 = 10 5 ⇒ x = 2 Therefore, the required value is x = 2 .