Solve the following equations and verify your answer : 17 2 - x 1 - 7 x + 5 x + 12 7 x - 1 = 8

We have: 17 2 - x 1 - 7 x + 5 x + 12 7 x - 1 = 8 ⇒ 17 2 - x 1 - 7 x - 5 x + 12 1 - 7 x = 8 ⇒ 17 2 - x - 5 x + 12 1 - 7 x = 8 ⇒ 34 - 17 x - 5 x - 60 1 - 7 x = 8 ⇒ - 22 x - 26 1 - 7 x = 8 By cross multiplication, we get ⇒ - 22 x - 26 = 8 1 - 7 x ⇒ - 22 x - 26 = 8 - 56 x Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ - 22 x + 56 x = 8 + 26 ⇒ 34 x = 34 ⇒ x = 1 To check: Substituting x = 1 in the given equation, we get LHS = 17 2 - 1 1 - 7 1 + 5 1 + 12 7 1 - 1 = 17 - 6 + 65 6 = - 17 + 65 6 = 48 6 = 8 RHS = 8 ∴ LHS = RHS Hence, x = 1 is a solution of the given equation.

Solve the following equations and verify your answer : 4 x - 3 10 x - 9 = 0 . 05 0 . 25

We have: 4 x - 3 10 x - 9 = 0 . 05 0 . 25 ⇒ 4 x - 3 10 x - 9 = 1 5 By cross multiplication, we get ⇒ 5 × ( 4 x - 3 ) = 1 × ( 10 x - 9 ) ⇒ 20 x - 15 = 10 x - 9 Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ 20 x - 10 x = - 9 + 15 ⇒ 10 x = 6 ⇒ x = 6 10 = 3 5 To check: Substituting x = 3 5 in the given equation, we get LHS = 4 3 5 - 3 10 3 5 - 9 = 12 - 15 30 - 45 = - 3 - 15 = 1 5 RHS = 0 . 05 0 . 25 = 1 5 ∴ LHS = RHS Hence, x = 3 5 is a solution of the given equation.

Solve the following equations and verify your answer : 3 2 x - 3 - 4 x = 1 3

We have: 3 2 x - 3 - 4 x = 1 3 By cross multiplication, we get ⇒ 3 × 3 = 2 x - 3 - 4 x ⇒ 9 = 2 x - 3 + 4 x Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ - 4 x - 2 x = - 3 - 9 ⇒ - 6 x = - 12 ⇒ x = 2 To check: Substituting x = 2 in the given equation, we get LHS = 3 2 2 - 3 - 4 2 = 3 4 - - 5 = 3 9 = 1 3 RHS = 1 3 ∴ LHS = RHS Hence, x = 2 is a solution of the given equation.

Solve the following equations and verify your answer : 5 x + 6 - 15 2 - x 3 x - 1 = 1 0

We have: 5 x + 6 - 15 2 - x 3 x - 1 = 10 ⇒ 5 x + 30 - 30 + 15 x 3 x - 1 = 10 ⇒ 20 x 3 x - 1 = 10 By cross multiplication, we get ⇒ 20 x = 10 × 3 x - 1 ⇒ 20 x = 30 x - 10 Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ 20 x - 30 x = - 10 ⇒ - 10 x = - 10 ⇒ x = 1 To check: Substituting x = 1 in the given equation, we get LHS = 5 1 + 6 - 15 2 - 1 3 1 - 1 = 35 - 15 2 = 20 2 = 10 RHS = 10 ∴ LHS = RHS Hence, x = 1 is a solution of the given equation.

Solve the following equations and verify your answer : x - 5 x - 3 = x - 4 x - 6

We have: x - 5 x - 3 = x - 4 x - 6 By cross multiplication, we get ⇒ x - 5 x - 6 = x - 4 x - 3 ⇒ x 2 - 6 x - 5 x + 30 = x 2 - 4 x - 3 x + 12 ⇒ x 2 - 11 x + 30 = x 2 - 7 x + 12 Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ x 2 - 11 x - x 2 + 7 x = 12 - 30 ⇒ - 4 x = - 18 ⇒ x = - 18 - 4 = 9 2 To check: Substituting x = 9 2 in the given equation, we get LHS = 9 2 - 5 9 2 - 3 = 9 - 10 9 - 6 = - 1 3 RHS = 9 2 - 4 9 2 - 6 = 9 - 8 9 - 12 = - 1 3 ∴ LHS = RHS Hence, x = 9 2 is a solution of the given equation.

Solve the following equations and verify your answer : 2 x + 9 = - 7 4 x - 3

We have: 2 x + 9 = - 7 4 x - 3 By cross multiplication, we get ⇒ 2 × 4 x - 3 = - 7 × x + 9 ⇒ 8 x - 6 = - 7 x - 63 Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ 8 x + 7 x = - 63 + 6 ⇒ 15 x = - 57 ⇒ x = - 57 15 = - 19 5 To check: Substituting x = - 19 5 in the given equation, we get LHS = 2 - 19 5 + 9 = 2 × 5 - 19 + 45 = 10 26 = 5 13 RHS = - 7 4 - 19 5 - 3 = - 7 × 5 - 76 - 15 = 35 91 = 5 13 ∴ LHS = RHS Hence, x = - 19 5 is a solution of the given equation.

Solve the following equations and verify your answer : 2 + 3 y 1 + 5 y = 4 3

We have: 2 + 3 y 1 + 5 y = 4 3 By cross multiplication, we get ⇒ 3 × 2 + 3 y = 4 × 1 + 5 y ⇒ 6 + 9 y = 4 + 20 y Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ 9 y - 20 y = 4 - 6 ⇒ - 11 y = - 2 ⇒ y = 2 11 To check: Substituting y = 2 11 in the given equation, we get LHS = 2 + 3 2 11 1 + 5 2 11 = 22 + 6 11 + 10 = 28 21 = 4 3 RHS = 4 3 ∴ LHS = RHS Hence, y = 2 11 is a solution of the given equation.

Solve the following equations and verify your answer : 3 x - 2 5 x - 1 = 3 x + 1 5 x + 2

We have: 3 x - 2 5 x - 1 = 3 x + 1 5 x + 2 By cross multiplication, we get ⇒ 3 x - 2 5 x + 2 = 3 x + 1 5 x - 1 ⇒ 15 x 2 - 10 x + 6 x - 4 = 15 x 2 + 5 x - 3 x - 1 ⇒ 15 x 2 - 4 x - 4 = 15 x 2 + 2 x - 1 Transposing variables to L.H.S. and constants to R.H.S., we get ⇒ 15 x 2 - 4 x - 15 x 2 - 2 x = - 1 + 4 ⇒ - 6 x = 3 ⇒ x = - 3 6 = - 1 2 To check: Substituting x = - 1 2 in the given equation, we get LHS = 3 - 1 2 - 2 5 - 1 2 - 1 = - 3 - 4 - 5 - 2 = - 7 - 7 = 1 RHS = 3 - 1 2 + 1 5 - 1 2 + 2 = - 3 + 2 - 5 + 4 = 1 ∴ LHS = RHS Hence, x = - 1 2 is a solution of the given equation.

E M B I B E

Step by Step>Chapter 8 - Linear Equations in One Variable>EXERCISE 8.2

Jagat Jitendra Matto Solutions for Chapter: Linear Equations in One Variable, Exercise 2: EXERCISE 8.2

Author:Jagat Jitendra Matto

Jagat Jitendra Matto Mathematics Solutions for Exercise - Jagat Jitendra Matto Solutions for Chapter: Linear Equations in One Variable, Exercise 2: EXERCISE 8.2

Attempt the practice questions on Chapter 8: Linear Equations in One Variable, Exercise 2: EXERCISE 8.2 with hints and solutions to strengthen your understanding. Step by Step solutions are prepared by Experienced Embibe Experts.