Solve the following equation and check your result: 4 5 x + 12 = - 7

The given equation is, 4 5 x + 12 = - 7 Now, subtract by 12 from both the sides, we get, ⇒ 4 5 x + 12 - 12 = - 7 - 12 ⇒ 4 5 x = - 19 Now, multiplying by 5 4 from both the sides of the equation, we get, ⇒ 4 5 x × 5 4 = - 19 × 5 4 ⇒ x = - 95 4 Check: Substitute the value of x in LHS. 4 5 × - 95 4 + 12 = - 19 + 12 = - 7 = RHS Therefore, LHS = RHS Hence, verified.

Solve the equation and check your result: p 5 + 8 = - 7

The given equation is, p 5 + 8 = - 7 Now, subtract by 8 from both the side, we get, ⇒ p 5 + 8 - 8 = - 7 - 8 ⇒ p 5 = - 15 Now, multiplying by 5 from both the sides of the equation, we get, ⇒ p 5 × 5 = - 15 × 5 ⇒ p = - 75 Check: Substitute the value of x in LHS. - 75 5 + 8 = - 15 + 8 = - 7 = RHS Therefore, LHS = RHS Hence, verified.

Solve the following equation and check your result: x 2 + x 3 + x = 18

The given equation is, x 2 + x 3 + x = 18 Now, the LCM of 2 , 3 , 1 = 6 ⇒ 3 x + 2 x + 6 x 6 = 18 ⇒ 11 x 6 = 18 ⇒ 11 x = 18 × 6 ⇒ x = 108 11 Checking: Substitute the value of x = 108 11 in LHS ⇒ 108 2 × 11 + 108 3 × 11 + 108 11 = 324 + 216 + 648 66 = 1188 66 = 18 = RHS Therefore, LHS = RHS Hence, verified.

Solve the following equation and check the result. x + 2 6 - ( 11 - x 3 - 1 4 ) = 3 x - 4 12

Given equation is x + 2 6 - ( 11 - x 3 - 1 4 ) = 3 x - 4 12 x + 2 6 - 11 - x 3 + 1 4 = 3 x - 4 12 2 ( x + 2 ) - 4 ( 11 - x ) + 3 12 = 3 x - 4 12 2 x + 4 - 44 + 4 x + 3 12 = 3 x - 4 12 6 x - 37 12 = 3 x - 4 12 6 x - 37 = 3 x - 4 6 x - 3 x = - 4 + 37 3 x = 33 x = 33 3 x = 11 Hence, the value of x is 11 . Checking: Substitute the value of x in L.H.S 11 + 2 6 - ( 11 - 11 3 - 1 4 ) = 13 6 + 1 4 = 26 + 3 12 = 29 12 Substitute the value of x in R.H.S 3 × 11 - 4 12 = 33 - 4 12 = 29 12 L.H.S=R.H.S Hence, the result is verified.

E M B I B E

New Global Mathematics - 7>Chapter 7 - Linear Equation in One Variable>EXERCISE 7.1

Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Linear Equation in One Variable, Exercise 1: EXERCISE 7.1

Author:Dr Pooranmal & Ashfaq A Khan

Dr Pooranmal Mathematics Solutions for Exercise - Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Linear Equation in One Variable, Exercise 1: EXERCISE 7.1

Attempt the practice questions on Chapter 7: Linear Equation in One Variable, Exercise 1: EXERCISE 7.1 with hints and solutions to strengthen your understanding. New Global Mathematics - 7 solutions are prepared by Experienced Embibe Experts.