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

Given linear equation is: x - 5 3 = 3 x - 1 5 + 2 By simplifying, we get x - 5 3 - 3 x - 1 5 = 2 5 x - 5 - 3 3 x - 1 15 = 2 5 x - 25 - 9 x + 3 = 30 - 4 x - 22 = 30 Adding 22 on both sides, we get - 4 x = 52 Dividing - 4 on both sides, we get, x = - 13 Hence, x = - 13 is the solution to equation x - 5 3 = 3 x - 1 5 + 2 Verification: Substituting x = - 13 in left-hand side of the equation LHS ⇒ x - 5 3 = - 13 - 5 3 = - 6 Substituting x = - 13 in the right-hand side of the equation, we get RHS ⇒ 3 x - 1 5 + 2 = 3 × - 13 - 1 + 10 5 = - 30 5 = - 6 Since LHS = RHS Hence, verified.

Solve the following equations and verify the solution: x + 2 3 - x - 3 4 = 5 - x - 1 2

Given linear equation is: x + 2 3 - x - 3 4 = 5 - x - 1 2 By simplifying we get, 4 x + 2 - 3 x - 3 12 = 5 - x - 1 2 4 x + 8 - 3 x + 9 12 = 5 - x - 1 2 x + 17 12 = 5 - x - 1 2 Shifting the variable to one side by adding x - 1 2 on both sides, we get x + 17 12 + x - 1 2 = 5 x + 17 + 6 x - 1 12 = 5 x + 17 + 6 x - 6 = 60 7 x + 11 = 60 Subtracting 11 on both sides, we get 7 x = 49 Dividing both sides by 7 , we get x = 7 Hence, x = 7 is the solution to equation x + 2 3 - x - 3 4 = 5 - x - 1 2 Verification: Substituting x = 7 in left-hand side of the equation LHS ⇒ x + 2 3 - x - 3 4 = 9 3 - 4 4 = 3 - 1 = 2 Substituting x = 7 in right-hand side of the equation RHS ⇒ 5 - x - 1 2 = 5 - 6 2 = 2 Since LHS = RHS Hence, verified.

Solve the following equations and verify the solution: 3 4 y + 2 y = 1 2 + 4 y - 3

Given linear equation is: 3 4 y + 2 y = 1 2 + 4 y - 3 By simplifying the equation, we get, 11 y 4 = 4 y - 5 2 Subtracting 4 y on both sides, we get 11 y 4 - 4 y = - 5 2 11 y - 16 y 4 = - 5 2 - 5 y 4 = - 5 2 Dividing both sides by - 5 4 , we get y = - 5 2 × - 4 5 = 2 Hence, y = 2 is the solution to equation 3 4 y + 2 y = 1 2 + 4 y - 3 Verification: Substituting y = 2 in left-hand side of the equation LHS ⇒ 3 4 y + 2 y = 3 4 × 2 + 2 × 2 = 3 2 + 4 = 11 2 Substituting y = 2 in right-hand side of the equation RHS ⇒ 1 2 + 4 y - 3 = 1 2 + 4 × 2 - 3 = 1 2 + 5 = 11 2 Since LHS = RHS Hence, verified.

E M B I B E

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2

Rupesh Pathak and Vaishali Bhatia Solutions for Chapter: Linear Equations in one Variable, Exercise 4: Exercise 7.2

Author:Rupesh Pathak & Vaishali Bhatia

Rupesh Pathak Mathematics Solutions for Exercise - Rupesh Pathak and Vaishali Bhatia Solutions for Chapter: Linear Equations in one Variable, Exercise 4: Exercise 7.2

Attempt the practice questions on Chapter 7: Linear Equations in one Variable, Exercise 4: Exercise 7.2 with hints and solutions to strengthen your understanding. 'I Did It' Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Rupesh Pathak and Vaishali Bhatia Solutions for Chapter: Linear Equations in one Variable, Exercise 4: Exercise 7.2 with Hints & Solutions

MEDIUM

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$\frac{x - 5}{3} = \frac{3 x - 1}{5} + 2$

MEDIUM

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$\frac{x + 1}{4} + 0.25 = x$

HARD

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$\frac{x + 2}{3} - \frac{x - 3}{4} = 5 - \frac{x - 1}{2}$

HARD

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$(0.5 y - 2) - 1 = 1 - (0.5 y - 2)$

HARD

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$5 x + \frac{7}{2} = \frac{3}{2} x - 14$

HARD

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$5 y - \frac{y + 1}{3} = 6 y + \frac{1}{5}$

HARD

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 1

Solve the following equations and verify the solution:

$\frac{3}{4} y + 2 y = \frac{1}{2} + 4 y - 3$

MEDIUM

8th CBSE

IMPORTANT

'I Did It' Mathematics>Chapter 7 - Linear Equations in one Variable>Exercise 7.2>Q 2

A mother is $24$ years older than her son. After $4$ years, the age of the mother will be three times the age of the son. Find the present ages of mother and son.

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