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.
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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 RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev
 (v) y = 2 is the root of y − 3 = 2y − 5
 (vi) x = 8 is the root of RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

 

Answer 1:

(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 RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev
  Now, substituting x = 4 in place of 'x' in the given equation RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

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

LHS = RHS
  Hence, x = 4 is the root of  RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

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

Now, substituting x = 8 in place of 'x' in the given equation RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

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

4 + 7 = 11  
   11 = 11
  LHS = RHS
Hence, x = 8 is the root of RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Question 2:

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

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

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

Answer 2:

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

Here, LHS =  RD Sharma Solutions - Ex - 8.1, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev  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.

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

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

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.

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

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

 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.

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

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

 

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.

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

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.

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

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

 

 

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