RS Aggarwal Solutions Class 7 Maths Chapter 7 Ex 7.1 (Updated For 2024)

RS Aggarwal Solutions Class 7 Maths Chapter 7 Ex 7.1: Ace your Class 7 Maths exams with RS Aggarwal Solutions Class 7 Maths. Be it your class assignments or exam preparations, RS Aggarwal Solutions Class 7 Maths Chapter 7 Ex 7.1 has got your back, All the solutions are prepared by subject matter experts and are accurate as well as credible.

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RS Aggarwal Solutions Class 7 Maths Chapter 7 Ex 7.1

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Ex 7A Solutions

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = 53
Check:
L.H.S. = 3x – 5
= 3 x 53 – 5
= 5 – 5
= 0
= R.H.S.
Hence x = 53

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = 714 = 12
Check : L.H.S. = 2(x – 2) + 3 (4x -1)

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:

Question 8.
Solution:

L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

RS Aggarwal Solutions Class 7 Maths Chapter 7 Ex 7.1 – Overview

Linear Equation in One Variable is defined as an equation that consists of one variable having order 1 at the max. It is usually represented in the form of “Qx + r = 0”, where ‘x’ represents the variable value. This equation will have the same values on the right and the left sides giving only one solution. Let us understand the few forms of Linear Equation in One Variable below –

7x = 1

77x – 1= 0

5x + 7 = -24

‘Q’ and ‘R’ are the real numbers, also both these  ‘Q’ and ‘R’ numbers are not equivalent to zero.

Hence, the formula of the linear equation – Having only one variable can be represented as Qx + R = 0.

You have to follow the basic three steps to solve any “Linear Equations having One Variable”.

1. Change the fractional units by making use of the LCM method.
2. Simplify the left and the right sides of the given equation.
3. Extract the variable.
4. Finally, just verify your solution.

Steps To Solve The Linear Equations

Let us consider an equation – 3x – 10 = -5x + 17 

1. Bring all the variables to one side – either left or either right, i.e., transposing. During the happening of this procedure, the general operation that we do on the operand is reversed.

   3x – 10 + 5x = 17 

   8x – 10 = 17 

2.  Similarly, transpose all the terms having represented the constant values to the right side of the equation. It will yield the following equation 

   8x = 17 + 10 

    8x = 27

3. Divide both sides of the equation by the number – “8”. By doing so, it will yield the following equation –

{ (8x / 8) = ( 27 / 8 ) } 

x = (27 / 8) 

Hence, you have found the value of the variable ‘x’, so let us then put its value in the main equation. By doing so, it will yield the following final equation as stated below –

After putting the values, it becomes –

3 (27 / 8) – 10 = -5 (27 / 8) + 17 

 (81 – 80) / 8 = (-135 + 136 ) / 8

(1 / 8) = (1 / 8) 

Hence, we can see that both sides of the equation are the same.

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