# NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.3

NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.3 (Ex. 4.3) Simple Equations in English and Hindi Medium updated for academic session 2020-2021. All the solutions and other study material are free to use without any login or password.

In class 7 math exercise 4.3, we will learn the way to solve the linear equations in one variable. All the questions in ex. 4.4 are easy to understand and simple to solve.## Class 7 Maths Chapter 4 Exercise 4.3 Solution

Class: 7 | Mathematics |

Chapter: 4 | Simple Equations |

Exercise: 4.3 | English and Hindi Medium Solution |

### CBSE NCERT Class 7 Maths Chapter 4 Exercise 4.3 Solution in Hindi and English Medium

### Class 7 Maths Chapter 4 Exercise 4.3 Solution in Videos

##### From Solution to Equation

We can get equation from a solution. this is reverse order to find a equation from its solution.

Equation → Solution (normal path)

Solution → Equation (reverse path)

Start with x = 5

Multiply both sides by 4,

4x = 20

Subtract 3 from both sides, we get:

4x – 3 = 17

This has resulted in an equation.

Note: (i) Is it not nice that not only can you solve an equation, but you can make equations.

(ii) From a given equation, you get one solution; but given a solution, you can make many equations.

##### Nine added to thrice a whole number gives 45. Find the number.

Let the required number be x. Then, 3x + 9 = 45

Or, 3x = 45 – 9

Or, 3x = 36

Or, x = = 12.

Hence, the required number is 12.

##### Solve: 12p – 5 = 25

Adding 5 on both sides of the equation,

12p – 5 + 5 = 25 + 5 or 12p = 30

Dividing both sides by 12,

12p/12 = 30/12

or, p = 5/2

Check Putting p = 5/2 in the LHS

LHS = 12 (5/2) – 5

or, 30 – 5 = 25 equal to RHS

##### The sum of two consecutive multiples of 3 is 69. Find them.

Let the two consecutive multiples of 3 be 3x and 3 (x + 1).

Then, 3x + 3 (x + 1) = 69

Or, 3x + 3x + 3 = 69

Or, 6x = 66

Or, x = 66/6 = 11

Hence, the required multiples of 3 are 3 x 11 and 3 x 12, that is, 33 and 36.

##### The length of a rectangular plot exceeds its breadth by 5 metres. If the perimeter of the plot is 142 metres. Find the dimensions of the plot.

Let the breadth of the plot be x metres.

Then, the length = (x + 5) metres.

So, perimeter = 2 (length + breadth)

= 2 (x + 5 + x) metres

= (4x + 10) metres.

4x + 10 = 142

Or, 4x = 132

Or, x = = 33.

Thus, breadth = 33 metres.

And, length = (33 + 5) metres = 38 metres.

Hence, length = 38 metres and breadth = 33 metres

##### How do you solve linear equations in one variable word problems?

Steps involved in solving a linear equation word problem:

(i) Denote the unknown by the variables as x, y, …….

(ii) Translate the problem to the language of mathematics or mathematical statements.

(iii) Form the linear equation in one variable using the conditions given in the problems.

(iv) Solve the equation for the unknown.

##### What is the graph of linear equation in one variable?

The graph of a linear equation in one variable x forms a vertical line parallel to the y-axis and vice-versa, whereas the graph of a linear equation in two variables x and y forms a straight line.

##### The numerator of a fraction is 4 less than the denominator. If 1 is added to both its numerator and denominator, it becomes 1/2. Find the fraction.

Let the denominator of the required fraction be x

So, {(x – 4) + 1}/(x+1) = ½

Or, (x – 3)/ (x + 1) = ½

Or, 2(x – 3) = x + 1

Or, 2x – 6 = x + 1

Or, 2x – x = 1 + 6

Or, x = 7

So, Denominator is 7 and numerator = 7 – 4 = 3

Hence, the number is 3/7.