# NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.2

NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.2 (Ex. 4.2) Simple Equations in Hindi and English Medium free to use online or download to use offline. Solution are based on latest CBSE books for class 7 issued for session 2020-2021.

Step by step solutions of exercise 4.2 of class 7 are given here in simplified format, so that student can understand easily. Videos Solution are also given with the PDF solution.## Class 7 Maths Chapter 4 Exercise 4.2 Solution

Class: 7 | Mathematics |

Chapter: 4 | Simple Equations |

Exercise: 4.2 | PDF and Videos Solution |

### CBSE NCERT Class 7 Maths Chapter 4 Exercise 4.2 Solution in Hindi and English Medium

### Class 7 Maths Chapter 4 Exercise 4.2 Solution in Videos

#### Linear Equation in One Variable

The linear equations in one variable is an equation which is expressed in the form of ax + b = 0, where a and b are two integers, and x is a variable and has only one solution.

For example, 2x + 3 = 8 is a linear equation having a single variable in it.

##### Solve: 2y + 11/4 = (1/3) y + 2

We have:

2y + 11/4 = (1/3) y + 2

Or, 2y – (1/3) y = 2 – 11/4

Or, (6y – y)/3 = (8 – 11)/4

Or, 5y/3 = -3/4

Or, y = – 9/20

Thus, y = – 9/20 is a solution of the given equation.

Substituting y = -9/20 in the given equation. we get,

LHS = 2 x (-9/20) + 11/4 = -9/10 + 11/4 = (-18 + 55)/20 = 37/20

RHS = 1/3 x (-9/20) + 2 = -3/20 + 2/1 = (-3 + 40)/20 = 37/20

So, LHS = RHS

Hence, y = -9/20 is a solution of the given equation.

##### Simple Equations

As we have seen, while solving equations one commonly used operation is adding or

subtracting the same number on both sides of the equation.

#### Transposition:

Any term of an equation may be taken from one side to the other with a change in its sign. This does not affect the equality of the statement. This process is called transposition.

Transposing a number (i.e., changing the side of the number) is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed.

##### Solve: (5x – 4)/8 – (x – 3)/ 5 = (x + 6)/4.

We have:

(5x – 4)/8 – (x – 3)/ 5 = (x + 6)/4

Multiplying both sides by 40, the LCM of 8, 5 and 4, we get

5 (5x – 4) – 8 (x – 3) = 10(x + 6)

Or, 25x – 20 – 8x + 24 = 10x + 60

Or, 17x + 4 = 10x + 60 [transposing 10x to LHS and 4 to RHS]

Or, 17x – 10x = 60 – 4

Or, 7x = 56

Or, x = 56/7 = 8

Thus, x = 8 is a solution of the given equation.

Check: Substituting x = 8 in the given equation, we get,

LHS = (5 x 8 – 4)/8 – (8 – 3)/ 5 = 36/8 – 1 = 28/8 = 7/2

RHS = (8 + 6)/4 = 14/4 = 7/2

So, LHS = RHS

Hence, x = 8 is a solution of the given equation.

##### Why do we use linear equations?

Linear equations are an important tool in science and many everyday applications. They allow scientist to describe relationships between two variables in the physical world, make predictions, calculate rates, and make conversions, among other things. Graphing linear equations helps make trends visible.

##### How do you know if a system is linear?

If the relationship between y and x is linear (straight line) and crossing through origin then the system is linear. If you find any time t at which the system is not linear then the system is non-linear.

##### Solve: 0.3x + 0.4 = 0.28x + 1.16

We have:

0.3 x + 0.4 = 0.28x + 1.16

Or, 0.3 x – 0.28x = 1.16 – 0.4 [by transposition]

Or, 0.02x = 0.76

x = 0.72/0.02 = 72/2 = 38.

So, x = 38 is a solution of the given equation.