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# RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

## RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

## Class 7 : RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 1:

Verify by substitution that:
(i) x = 4 is the root of 3x − 5 = 7
(ii) x = 3 is the root of 5 + 3x = 14
(iii) x = 2 is the root of 3x − 2 = 8x − 12
(iv) x = 4 is the root of
(v) y = 2 is the root of y − 3 = 2y − 5
(vi) x = 8 is the root of

(i) x = 4 is the root of 3x − 5 = 7.
Now, substituting x = 4 in place of 'x' in the given equation 3x − 5 = 7,
3(4) − 5 = 7
12 − 5 = 7
7 = 7
LHS = RHS
Hence, x = 4 is the root of 3x − 5 = 7.

(ii) x = 3 is the root of 5 + 3x = 14.
Now, substituting x = 3 in place of 'x' in the given equation 5 + 3x = 14,
5 + 3(3) = 14
5 + 9 = 14
14 = 14
LHS = RHS
Hence, x = 3 is the root of 5 + 3x = 14.

(iii) x = 2 is the root of 3x − 2 = 8x − 12.
Now, substituting x = 2 in place of 'x' in the given equation 3x − 2 = 8x − 12,
3(2) − 2 = 8(2) − 12
6 − 2 = 16 − 12
4 = 4
LHS = RHS
Hence, x = 2 is the root of 3x − 2 = 8x − 12.

(iv) x = 4 is the root of
Now, substituting x = 4 in place of 'x' in the given equation

LHS = RHS
Hence, x = 4 is the root of

(v) y = 2 is the root of y − 3 = 2y − 5.
Now, substituting y = 2 in place of 'y' in the given equation y − 3 = 2y − 5,
2 − 3 = 2(2) − 5
−1 = 4 − 5
−1 = −1
LHS = RHS
Hence, = 2 is the root of y − 3 = 2y − 5.

(vi) x = 8 is the root of

Now, substituting x = 8 in place of 'x' in the given equation

4 + 7 = 11
11 = 11
LHS = RHS
Hence, x = 8 is the root of

#### Question 2:

Solve each of the following equations by trial-and-error method:

(i) x + 3 = 12
(ii) x − 7 = 10
(iii) 4x = 28

(i) x + 3 = 12
Here, LHS = x + 3 and RHS = 12.

Therefore, if x = 9, LHS = RHS.
Hence, x = 9 is the solution to this equation.

(ii) x − 7 = 10
Here, LHS = x −7 and RHS =10.

Therefore, if x = 17, LHS = RHS.
Hence, x = 17 is the solution to this equation.

(iii) 4x = 28
Here, LHS = 4x and RHS = 28.

Therefore, if x = 7, LHS = RHS.
Hence, x = 7 is the solution to this equation.

Here, LHS =    and RHS = 11.

Since RHS is a natural number, x/2 must also be a natural number, so we must substitute values of x that are multiples of 2.

Therefore, if x = 8, LHS = RHS.
Hence, x = 8 is the solution to this equation.

(v) 2x + 4 = 3x
Here, LHS = 2x + 4 and RHS = 3x.

Therefore, if x = 4, LHS = RHS.
Hence, x = 4 is the solution to this equation.

(vi) x/4 = 12
Here, LHS =x/4  and RHS = 12.
Since RHS is a natural number, x/4 must also be a natural number, so we must substitute values of x that are multiples of 4.

Therefore, if x = 48, LHS = RHS.
Hence, x = 48 is the solution to this equation.

(vii) 15/x = 3

Here, LHS =15/x  and RHS = 3.
Since RHS is a natural number, 15/x must also be a natural number, so we must substitute values of x that are factors of 15.

Therefore, if x = 5, LHS = RHS.
Hence, x = 5 is the solution to this equation.

(viii) x/18= 20
Here, LHS =x/18  and RHS = 20.
Since RHS is a natural number, x/18 must also be a natural number, so we must substitute values of x that are multiples of 18.

Therefore, if x = 360, LHS = RHS.
Hence, x = 360 is the solution to this equation.

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