RS Aggarwal Solutions: Linear Equation in One Variable Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : RS Aggarwal Solutions: Linear Equation in One Variable Class 6 Notes | EduRev

 Page 1


Points to Remember :
Equation. A statement of equality which involves one or more variables is called ab equation.
Linear equation. An equation in which the highest power of the variable involved is 1, is called a
linear equation.
Solution of an equation. A number which when substituted for the variable in an equation, makes
L.H.S. = R.H.S., is said to satisfy the equation and is called a solution or root of the equation.
(i) x – 7 = 14 (ii) 2 y = 18
(iii) 11 + 3 x = 17     (iv) 2 x – 3 = 13
(v) 12 y – 30 = 6 (vi) 
2
3
8
z
Sol. (i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is
17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number y is 6.
(vi) Quotient of twice the number z and 3 is
8.
Q. 3. Verify by substitution that :
(i) The root of 3 x – 5 = 7 is x = 4.
(ii) The root of 3 + 2 x = 9 is x = 3.
(iii) The root of 5 x – 8 = 2 x –2 is x = 2.
(iv) The root of 8 – 7 y = 1 is y = 1.
(v) The root of 
z
7
8 is z = 56.
Sol. (i) The given equation is 3 x – 5 = 7
Substituting x = 4, we get
     L.H.S. = 3 x – 5
= 3 × 4 – 5
= 12 – 5 = 7
= R.H.S.
( ) EXERCISE 9 A
Q. 1. Write each of the following statements
as an equation :
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice a number is
16.
(vi) If 12 is subtracted from a number, the
result is 24.
(vii) Twice a number subtracted from 19 is
11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the
number.
Sol. Let x be the given number, then
(i) 5 x = 40 (ii) x + 8 = 15
(iii) 25 – x = 7 (iv) x – 5 = 3
(v) 3 x – 5 = 16        (vi) x – 12 = 24
(vii) 19 – 2 x = 11 (viii) 
x
8
7
(ix) 4 x – 3 = 17 (x) 6 x = x + 5
Q. 2. Write a statement for each of the
equations, given below :
Page 2


Points to Remember :
Equation. A statement of equality which involves one or more variables is called ab equation.
Linear equation. An equation in which the highest power of the variable involved is 1, is called a
linear equation.
Solution of an equation. A number which when substituted for the variable in an equation, makes
L.H.S. = R.H.S., is said to satisfy the equation and is called a solution or root of the equation.
(i) x – 7 = 14 (ii) 2 y = 18
(iii) 11 + 3 x = 17     (iv) 2 x – 3 = 13
(v) 12 y – 30 = 6 (vi) 
2
3
8
z
Sol. (i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is
17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number y is 6.
(vi) Quotient of twice the number z and 3 is
8.
Q. 3. Verify by substitution that :
(i) The root of 3 x – 5 = 7 is x = 4.
(ii) The root of 3 + 2 x = 9 is x = 3.
(iii) The root of 5 x – 8 = 2 x –2 is x = 2.
(iv) The root of 8 – 7 y = 1 is y = 1.
(v) The root of 
z
7
8 is z = 56.
Sol. (i) The given equation is 3 x – 5 = 7
Substituting x = 4, we get
     L.H.S. = 3 x – 5
= 3 × 4 – 5
= 12 – 5 = 7
= R.H.S.
( ) EXERCISE 9 A
Q. 1. Write each of the following statements
as an equation :
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice a number is
16.
(vi) If 12 is subtracted from a number, the
result is 24.
(vii) Twice a number subtracted from 19 is
11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the
number.
Sol. Let x be the given number, then
(i) 5 x = 40 (ii) x + 8 = 15
(iii) 25 – x = 7 (iv) x – 5 = 3
(v) 3 x – 5 = 16        (vi) x – 12 = 24
(vii) 19 – 2 x = 11 (viii) 
x
8
7
(ix) 4 x – 3 = 17 (x) 6 x = x + 5
Q. 2. Write a statement for each of the
equations, given below :
 It is verified that x = 4 is the root of
the given equation.
(ii) The given equation is 3 + 2 x = 9
Substituting x = 3, we get
L.H.S. = 3 + 2 x
 = 3 + 2 × 3
 = 3 + 6
 = 9
 = R.H.S.
 It is verified that x = 3 is the root of
the given equation.
(iii) The given equation is 5 x – 8 = 2 x – 2
Substituting x = 2, we get
L.H.S. = 5 x – 8      R.H.S. = 2 x – 2
= 5 × 2 – 8      = 2 × 2 – 2
= 10 – 8      = 4 – 2
= 2      = 2
 L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the
root of the given equation.
(iv) The given equation is 8 – 7 y = 1
Substituting y = 1, we get
L.H.S. = 8 – 7 y
 = 8 – 7 × 1
 = 8 – 7
 = 1
 = R.H.S.
Hence, it verified that y = 1 is the root
of the given equation.
(v) The given equation is 
z
7
8
Substituting the value of z = 56, we get
L. H.S.=
z
7
56
7
= 8
= R.H.S.
Hence, it is verified that z = 56 is the
root of the given equation.
Q. 4. Solve each of the following equations
by trial and error method :
(i) y + 9 = 13 (ii) x – 7 = 10
(iii) 4 x = 28 (iv) 3 y = 36
(v) 11 + x = 19 (vi) 
x
3
4
(vii) 2 x – 3 = 9 (viii) 
1
2
7 11 x
(ix) 2 y + 4 = 3 y (x) z – 3 = 2z – 5
Sol. (i) The given equation is y + 9 = 13
We try several values of y and find L.H.S.
and the R.H.S. and stop when L.H.S. =
R.H.S.
y L.H.S.    R.H.S.
2 2 + 9 = 11   13
3 3 + 9 = 12   13
4 4 + 9 = 13   13
 When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given
equation.
(ii) The given equation is x – 7 = 10
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x   L.H.S.                 R.H.S.
14  14 – 7 = 7     10
15   15 – 7 = 8     10
16   16 – 7 = 9        10
17   17 – 7 = 10        10
 When x = 17, we have L.H.S. =
R.H.S.
So x = 17 is the solution of the given
equation.
(iii) The given equation is 4x = 28
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
Page 3


Points to Remember :
Equation. A statement of equality which involves one or more variables is called ab equation.
Linear equation. An equation in which the highest power of the variable involved is 1, is called a
linear equation.
Solution of an equation. A number which when substituted for the variable in an equation, makes
L.H.S. = R.H.S., is said to satisfy the equation and is called a solution or root of the equation.
(i) x – 7 = 14 (ii) 2 y = 18
(iii) 11 + 3 x = 17     (iv) 2 x – 3 = 13
(v) 12 y – 30 = 6 (vi) 
2
3
8
z
Sol. (i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is
17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number y is 6.
(vi) Quotient of twice the number z and 3 is
8.
Q. 3. Verify by substitution that :
(i) The root of 3 x – 5 = 7 is x = 4.
(ii) The root of 3 + 2 x = 9 is x = 3.
(iii) The root of 5 x – 8 = 2 x –2 is x = 2.
(iv) The root of 8 – 7 y = 1 is y = 1.
(v) The root of 
z
7
8 is z = 56.
Sol. (i) The given equation is 3 x – 5 = 7
Substituting x = 4, we get
     L.H.S. = 3 x – 5
= 3 × 4 – 5
= 12 – 5 = 7
= R.H.S.
( ) EXERCISE 9 A
Q. 1. Write each of the following statements
as an equation :
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice a number is
16.
(vi) If 12 is subtracted from a number, the
result is 24.
(vii) Twice a number subtracted from 19 is
11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the
number.
Sol. Let x be the given number, then
(i) 5 x = 40 (ii) x + 8 = 15
(iii) 25 – x = 7 (iv) x – 5 = 3
(v) 3 x – 5 = 16        (vi) x – 12 = 24
(vii) 19 – 2 x = 11 (viii) 
x
8
7
(ix) 4 x – 3 = 17 (x) 6 x = x + 5
Q. 2. Write a statement for each of the
equations, given below :
 It is verified that x = 4 is the root of
the given equation.
(ii) The given equation is 3 + 2 x = 9
Substituting x = 3, we get
L.H.S. = 3 + 2 x
 = 3 + 2 × 3
 = 3 + 6
 = 9
 = R.H.S.
 It is verified that x = 3 is the root of
the given equation.
(iii) The given equation is 5 x – 8 = 2 x – 2
Substituting x = 2, we get
L.H.S. = 5 x – 8      R.H.S. = 2 x – 2
= 5 × 2 – 8      = 2 × 2 – 2
= 10 – 8      = 4 – 2
= 2      = 2
 L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the
root of the given equation.
(iv) The given equation is 8 – 7 y = 1
Substituting y = 1, we get
L.H.S. = 8 – 7 y
 = 8 – 7 × 1
 = 8 – 7
 = 1
 = R.H.S.
Hence, it verified that y = 1 is the root
of the given equation.
(v) The given equation is 
z
7
8
Substituting the value of z = 56, we get
L. H.S.=
z
7
56
7
= 8
= R.H.S.
Hence, it is verified that z = 56 is the
root of the given equation.
Q. 4. Solve each of the following equations
by trial and error method :
(i) y + 9 = 13 (ii) x – 7 = 10
(iii) 4 x = 28 (iv) 3 y = 36
(v) 11 + x = 19 (vi) 
x
3
4
(vii) 2 x – 3 = 9 (viii) 
1
2
7 11 x
(ix) 2 y + 4 = 3 y (x) z – 3 = 2z – 5
Sol. (i) The given equation is y + 9 = 13
We try several values of y and find L.H.S.
and the R.H.S. and stop when L.H.S. =
R.H.S.
y L.H.S.    R.H.S.
2 2 + 9 = 11   13
3 3 + 9 = 12   13
4 4 + 9 = 13   13
 When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given
equation.
(ii) The given equation is x – 7 = 10
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x   L.H.S.                 R.H.S.
14  14 – 7 = 7     10
15   15 – 7 = 8     10
16   16 – 7 = 9        10
17   17 – 7 = 10        10
 When x = 17, we have L.H.S. =
R.H.S.
So x = 17 is the solution of the given
equation.
(iii) The given equation is 4x = 28
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x L.H.S. R.H.S.
2 4 × 2 = 8 28
4 4 × 4 = 16 28
5 4 × 5 = 20 28
6 4 × 6 = 24 28
7 4 × 7 = 28 28
 When x = 7, we have L.H.S. = R.H.S.
So x = 7 is the solution of the given
equation.
(iv) The given equation is 3 y = 36
We guess and try several values of y to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  y L.H.S. R.H.S.
  7 3 × 7 = 21 36
  9 3 × 9 = 27 36
 10 3 × 10 = 30 36
 12 3 × 12 = 36 36
 When y = 12, we have L.H.S. = R.H.S.
So y = 12 is the solution of the given
equation.
(v) The given equation is 11 + x = 19
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3 11 + 3 = 14 19
  4 11 + 4 = 15 19
  6 11 + 6 = 17 19
  7 11 + 7 = 18 19
  8 11 + 8 = 19 19
 When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given
equation.
(vi) The given equation is 
x
3
4
.
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3
3
3
1
4
  6
6
3
2
4
  9
9
3
3
4
 12
12
3
4
4
 When x = 12, we have L.H.S. = R.H.S.
So, x = 12 is the solution of the given
equation.
(vii) The given equation is 2 x – 3 = 9
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3 2 × 3 – 3 = 3 9
  4 2 × 4 – 3 = 5 9
  5 2 × 5 – 3 = 7 9
  6 2 × 6 – 3 = 9 9
 When x = 6, we have L.H.S. = R.H.S.
So, x = 6 is the solution of the given
equation.
(viii) The given equation is 
1
2
7 11 x
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  2
1
2
2 7 8
11
  4
1
2
4 7 9
11
  6
1
2
6 7 10
11
  8
1
2
8 7 11
11
Page 4


Points to Remember :
Equation. A statement of equality which involves one or more variables is called ab equation.
Linear equation. An equation in which the highest power of the variable involved is 1, is called a
linear equation.
Solution of an equation. A number which when substituted for the variable in an equation, makes
L.H.S. = R.H.S., is said to satisfy the equation and is called a solution or root of the equation.
(i) x – 7 = 14 (ii) 2 y = 18
(iii) 11 + 3 x = 17     (iv) 2 x – 3 = 13
(v) 12 y – 30 = 6 (vi) 
2
3
8
z
Sol. (i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is
17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number y is 6.
(vi) Quotient of twice the number z and 3 is
8.
Q. 3. Verify by substitution that :
(i) The root of 3 x – 5 = 7 is x = 4.
(ii) The root of 3 + 2 x = 9 is x = 3.
(iii) The root of 5 x – 8 = 2 x –2 is x = 2.
(iv) The root of 8 – 7 y = 1 is y = 1.
(v) The root of 
z
7
8 is z = 56.
Sol. (i) The given equation is 3 x – 5 = 7
Substituting x = 4, we get
     L.H.S. = 3 x – 5
= 3 × 4 – 5
= 12 – 5 = 7
= R.H.S.
( ) EXERCISE 9 A
Q. 1. Write each of the following statements
as an equation :
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice a number is
16.
(vi) If 12 is subtracted from a number, the
result is 24.
(vii) Twice a number subtracted from 19 is
11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the
number.
Sol. Let x be the given number, then
(i) 5 x = 40 (ii) x + 8 = 15
(iii) 25 – x = 7 (iv) x – 5 = 3
(v) 3 x – 5 = 16        (vi) x – 12 = 24
(vii) 19 – 2 x = 11 (viii) 
x
8
7
(ix) 4 x – 3 = 17 (x) 6 x = x + 5
Q. 2. Write a statement for each of the
equations, given below :
 It is verified that x = 4 is the root of
the given equation.
(ii) The given equation is 3 + 2 x = 9
Substituting x = 3, we get
L.H.S. = 3 + 2 x
 = 3 + 2 × 3
 = 3 + 6
 = 9
 = R.H.S.
 It is verified that x = 3 is the root of
the given equation.
(iii) The given equation is 5 x – 8 = 2 x – 2
Substituting x = 2, we get
L.H.S. = 5 x – 8      R.H.S. = 2 x – 2
= 5 × 2 – 8      = 2 × 2 – 2
= 10 – 8      = 4 – 2
= 2      = 2
 L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the
root of the given equation.
(iv) The given equation is 8 – 7 y = 1
Substituting y = 1, we get
L.H.S. = 8 – 7 y
 = 8 – 7 × 1
 = 8 – 7
 = 1
 = R.H.S.
Hence, it verified that y = 1 is the root
of the given equation.
(v) The given equation is 
z
7
8
Substituting the value of z = 56, we get
L. H.S.=
z
7
56
7
= 8
= R.H.S.
Hence, it is verified that z = 56 is the
root of the given equation.
Q. 4. Solve each of the following equations
by trial and error method :
(i) y + 9 = 13 (ii) x – 7 = 10
(iii) 4 x = 28 (iv) 3 y = 36
(v) 11 + x = 19 (vi) 
x
3
4
(vii) 2 x – 3 = 9 (viii) 
1
2
7 11 x
(ix) 2 y + 4 = 3 y (x) z – 3 = 2z – 5
Sol. (i) The given equation is y + 9 = 13
We try several values of y and find L.H.S.
and the R.H.S. and stop when L.H.S. =
R.H.S.
y L.H.S.    R.H.S.
2 2 + 9 = 11   13
3 3 + 9 = 12   13
4 4 + 9 = 13   13
 When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given
equation.
(ii) The given equation is x – 7 = 10
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x   L.H.S.                 R.H.S.
14  14 – 7 = 7     10
15   15 – 7 = 8     10
16   16 – 7 = 9        10
17   17 – 7 = 10        10
 When x = 17, we have L.H.S. =
R.H.S.
So x = 17 is the solution of the given
equation.
(iii) The given equation is 4x = 28
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x L.H.S. R.H.S.
2 4 × 2 = 8 28
4 4 × 4 = 16 28
5 4 × 5 = 20 28
6 4 × 6 = 24 28
7 4 × 7 = 28 28
 When x = 7, we have L.H.S. = R.H.S.
So x = 7 is the solution of the given
equation.
(iv) The given equation is 3 y = 36
We guess and try several values of y to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  y L.H.S. R.H.S.
  7 3 × 7 = 21 36
  9 3 × 9 = 27 36
 10 3 × 10 = 30 36
 12 3 × 12 = 36 36
 When y = 12, we have L.H.S. = R.H.S.
So y = 12 is the solution of the given
equation.
(v) The given equation is 11 + x = 19
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3 11 + 3 = 14 19
  4 11 + 4 = 15 19
  6 11 + 6 = 17 19
  7 11 + 7 = 18 19
  8 11 + 8 = 19 19
 When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given
equation.
(vi) The given equation is 
x
3
4
.
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3
3
3
1
4
  6
6
3
2
4
  9
9
3
3
4
 12
12
3
4
4
 When x = 12, we have L.H.S. = R.H.S.
So, x = 12 is the solution of the given
equation.
(vii) The given equation is 2 x – 3 = 9
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3 2 × 3 – 3 = 3 9
  4 2 × 4 – 3 = 5 9
  5 2 × 5 – 3 = 7 9
  6 2 × 6 – 3 = 9 9
 When x = 6, we have L.H.S. = R.H.S.
So, x = 6 is the solution of the given
equation.
(viii) The given equation is 
1
2
7 11 x
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  2
1
2
2 7 8
11
  4
1
2
4 7 9
11
  6
1
2
6 7 10
11
  8
1
2
8 7 11
11
When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given
equation.
(ix) The given equation is 2 y + 4 = 3 y
We guess and try several values of y to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
 y L.H.S. R.H.S.
 1 2 × 1 + 4 = 6 3 × 1 = 3
 2 2 × 2 + 4 = 8 3 × 2 = 6
 3 2 × 3 + 4 = 10 3 × 3 = 9
 4 2 × 4 + 4 = 12 3 × 4 = 12
 When y = 4, we have L.H.S. = R.H.S.
So, y = 4 is the solution of the given
equation.
(x) The given equation is z – 3 = 2z – 5
We guess and try several values of z to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  z L.H.S. R.H.S.
  1 1 – 3 = – 2 2 × 1 – 5 = – 3
  2 2 – 3 = – 1 2 × 2 – 5 = – 1
 When z = 2, we have L.H.S. = R.H.S.
So, z = 2 is the solution of the given
equation.
Systematic Method for Solving an
Equation.
Equation : Rules for Solving a Linear
1. We can add the same number to both
sides of an equation.
2. We can subtract the same number from
both sides of an equation.
3. We can multiply both sides of an
equation by a same non-zero number.
4. We can divide both the sides of an
equation by a same non-zero number.
( ) EXERCISE 9 B
Solve each of the following equations
and verify the answer is each case :
Q. 1. x + 5 = 12
Sol. x + 5 = 12
 x + 5 – 5 = 12 – 5
(Subtracting 5 from both sides)
 x = 7
 x = 7 is the solution of the given
equation.
Check : Substituting x = 7 in the given
equation, we get
L.H.S. = 7 + 5 = 12 and R.H.S. = 12
 When x = 7, we have L.H.S. = R.H.S.
Q. 2. x + 3 = – 2
 x + 3 – 3 = – 2 – 3
(Subtracting 3 from both sides)
 x = – 5
 x = – 5 is the solution of the given
equation.
Check : Substituting x = – 5 in the given
equation, we get :
L.H.S. = – 5 + 3 = – 2 and R.H.S. = – 2
 When x = – 5,
we have L.H.S. = R.H.S.
Q. 3. x – 7 = 6
Sol. x – 7 = 6
 x – 7 + 7 = 6 + 7
(Adding 7 to both sides)
 x = 13
So, x = 13 is the solution of the given
equation.
Check : Substituting x = 13 in the given
equation, we get
L.H.S. = 13 – 7 = 6 and R.H.S. = 6
 When x = 13, we have L.H.S. = R.H.S.
Q. 4. x – 2 = – 5
Sol. x – 2 = – 5
 x – 2 + 2 = – 5 + 2
(Adding 2 on both sides)
 x = – 3
So, x = – 3 is the solution of the given
equation.
Page 5


Points to Remember :
Equation. A statement of equality which involves one or more variables is called ab equation.
Linear equation. An equation in which the highest power of the variable involved is 1, is called a
linear equation.
Solution of an equation. A number which when substituted for the variable in an equation, makes
L.H.S. = R.H.S., is said to satisfy the equation and is called a solution or root of the equation.
(i) x – 7 = 14 (ii) 2 y = 18
(iii) 11 + 3 x = 17     (iv) 2 x – 3 = 13
(v) 12 y – 30 = 6 (vi) 
2
3
8
z
Sol. (i) 7 less than from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 increased by thrice the number x is
17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number y is 6.
(vi) Quotient of twice the number z and 3 is
8.
Q. 3. Verify by substitution that :
(i) The root of 3 x – 5 = 7 is x = 4.
(ii) The root of 3 + 2 x = 9 is x = 3.
(iii) The root of 5 x – 8 = 2 x –2 is x = 2.
(iv) The root of 8 – 7 y = 1 is y = 1.
(v) The root of 
z
7
8 is z = 56.
Sol. (i) The given equation is 3 x – 5 = 7
Substituting x = 4, we get
     L.H.S. = 3 x – 5
= 3 × 4 – 5
= 12 – 5 = 7
= R.H.S.
( ) EXERCISE 9 A
Q. 1. Write each of the following statements
as an equation :
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice a number is
16.
(vi) If 12 is subtracted from a number, the
result is 24.
(vii) Twice a number subtracted from 19 is
11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the
number.
Sol. Let x be the given number, then
(i) 5 x = 40 (ii) x + 8 = 15
(iii) 25 – x = 7 (iv) x – 5 = 3
(v) 3 x – 5 = 16        (vi) x – 12 = 24
(vii) 19 – 2 x = 11 (viii) 
x
8
7
(ix) 4 x – 3 = 17 (x) 6 x = x + 5
Q. 2. Write a statement for each of the
equations, given below :
 It is verified that x = 4 is the root of
the given equation.
(ii) The given equation is 3 + 2 x = 9
Substituting x = 3, we get
L.H.S. = 3 + 2 x
 = 3 + 2 × 3
 = 3 + 6
 = 9
 = R.H.S.
 It is verified that x = 3 is the root of
the given equation.
(iii) The given equation is 5 x – 8 = 2 x – 2
Substituting x = 2, we get
L.H.S. = 5 x – 8      R.H.S. = 2 x – 2
= 5 × 2 – 8      = 2 × 2 – 2
= 10 – 8      = 4 – 2
= 2      = 2
 L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the
root of the given equation.
(iv) The given equation is 8 – 7 y = 1
Substituting y = 1, we get
L.H.S. = 8 – 7 y
 = 8 – 7 × 1
 = 8 – 7
 = 1
 = R.H.S.
Hence, it verified that y = 1 is the root
of the given equation.
(v) The given equation is 
z
7
8
Substituting the value of z = 56, we get
L. H.S.=
z
7
56
7
= 8
= R.H.S.
Hence, it is verified that z = 56 is the
root of the given equation.
Q. 4. Solve each of the following equations
by trial and error method :
(i) y + 9 = 13 (ii) x – 7 = 10
(iii) 4 x = 28 (iv) 3 y = 36
(v) 11 + x = 19 (vi) 
x
3
4
(vii) 2 x – 3 = 9 (viii) 
1
2
7 11 x
(ix) 2 y + 4 = 3 y (x) z – 3 = 2z – 5
Sol. (i) The given equation is y + 9 = 13
We try several values of y and find L.H.S.
and the R.H.S. and stop when L.H.S. =
R.H.S.
y L.H.S.    R.H.S.
2 2 + 9 = 11   13
3 3 + 9 = 12   13
4 4 + 9 = 13   13
 When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given
equation.
(ii) The given equation is x – 7 = 10
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x   L.H.S.                 R.H.S.
14  14 – 7 = 7     10
15   15 – 7 = 8     10
16   16 – 7 = 9        10
17   17 – 7 = 10        10
 When x = 17, we have L.H.S. =
R.H.S.
So x = 17 is the solution of the given
equation.
(iii) The given equation is 4x = 28
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
x L.H.S. R.H.S.
2 4 × 2 = 8 28
4 4 × 4 = 16 28
5 4 × 5 = 20 28
6 4 × 6 = 24 28
7 4 × 7 = 28 28
 When x = 7, we have L.H.S. = R.H.S.
So x = 7 is the solution of the given
equation.
(iv) The given equation is 3 y = 36
We guess and try several values of y to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  y L.H.S. R.H.S.
  7 3 × 7 = 21 36
  9 3 × 9 = 27 36
 10 3 × 10 = 30 36
 12 3 × 12 = 36 36
 When y = 12, we have L.H.S. = R.H.S.
So y = 12 is the solution of the given
equation.
(v) The given equation is 11 + x = 19
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3 11 + 3 = 14 19
  4 11 + 4 = 15 19
  6 11 + 6 = 17 19
  7 11 + 7 = 18 19
  8 11 + 8 = 19 19
 When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given
equation.
(vi) The given equation is 
x
3
4
.
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3
3
3
1
4
  6
6
3
2
4
  9
9
3
3
4
 12
12
3
4
4
 When x = 12, we have L.H.S. = R.H.S.
So, x = 12 is the solution of the given
equation.
(vii) The given equation is 2 x – 3 = 9
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  3 2 × 3 – 3 = 3 9
  4 2 × 4 – 3 = 5 9
  5 2 × 5 – 3 = 7 9
  6 2 × 6 – 3 = 9 9
 When x = 6, we have L.H.S. = R.H.S.
So, x = 6 is the solution of the given
equation.
(viii) The given equation is 
1
2
7 11 x
We guess and try several values of x to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  x L.H.S. R.H.S.
  2
1
2
2 7 8
11
  4
1
2
4 7 9
11
  6
1
2
6 7 10
11
  8
1
2
8 7 11
11
When x = 8, we have L.H.S. = R.H.S.
So, x = 8 is the solution of the given
equation.
(ix) The given equation is 2 y + 4 = 3 y
We guess and try several values of y to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
 y L.H.S. R.H.S.
 1 2 × 1 + 4 = 6 3 × 1 = 3
 2 2 × 2 + 4 = 8 3 × 2 = 6
 3 2 × 3 + 4 = 10 3 × 3 = 9
 4 2 × 4 + 4 = 12 3 × 4 = 12
 When y = 4, we have L.H.S. = R.H.S.
So, y = 4 is the solution of the given
equation.
(x) The given equation is z – 3 = 2z – 5
We guess and try several values of z to
find L.H.S. and R.H.S. and stop when
L.H.S. = R.H.S.
  z L.H.S. R.H.S.
  1 1 – 3 = – 2 2 × 1 – 5 = – 3
  2 2 – 3 = – 1 2 × 2 – 5 = – 1
 When z = 2, we have L.H.S. = R.H.S.
So, z = 2 is the solution of the given
equation.
Systematic Method for Solving an
Equation.
Equation : Rules for Solving a Linear
1. We can add the same number to both
sides of an equation.
2. We can subtract the same number from
both sides of an equation.
3. We can multiply both sides of an
equation by a same non-zero number.
4. We can divide both the sides of an
equation by a same non-zero number.
( ) EXERCISE 9 B
Solve each of the following equations
and verify the answer is each case :
Q. 1. x + 5 = 12
Sol. x + 5 = 12
 x + 5 – 5 = 12 – 5
(Subtracting 5 from both sides)
 x = 7
 x = 7 is the solution of the given
equation.
Check : Substituting x = 7 in the given
equation, we get
L.H.S. = 7 + 5 = 12 and R.H.S. = 12
 When x = 7, we have L.H.S. = R.H.S.
Q. 2. x + 3 = – 2
 x + 3 – 3 = – 2 – 3
(Subtracting 3 from both sides)
 x = – 5
 x = – 5 is the solution of the given
equation.
Check : Substituting x = – 5 in the given
equation, we get :
L.H.S. = – 5 + 3 = – 2 and R.H.S. = – 2
 When x = – 5,
we have L.H.S. = R.H.S.
Q. 3. x – 7 = 6
Sol. x – 7 = 6
 x – 7 + 7 = 6 + 7
(Adding 7 to both sides)
 x = 13
So, x = 13 is the solution of the given
equation.
Check : Substituting x = 13 in the given
equation, we get
L.H.S. = 13 – 7 = 6 and R.H.S. = 6
 When x = 13, we have L.H.S. = R.H.S.
Q. 4. x – 2 = – 5
Sol. x – 2 = – 5
 x – 2 + 2 = – 5 + 2
(Adding 2 on both sides)
 x = – 3
So, x = – 3 is the solution of the given
equation.
Check : Substituting x = – 3 in the given
equation, we get
L.H.S. = – 3 – 2 = – 5 and R.H.S. = – 5
 When x = – 3, we have
L.H.S. = R.H.S.
Q. 5. 3 x – 5 = 13
Sol. 3 x – 5 = 13
 3 x – 5 + 5 = 13 + 5
(Adding 5 on both sides)
 3 x = 18
3
3
18
3
x
(Dividing both sides by 3)
 x = 6
 x = 6 is the solution of the given
equation.
Check : Substituting x = 6 in the given
equation, we get
L.H.S. = 3 × 6 – 5 = 18 – 5 = 13 and
R.H.S. = 13
 When x = 6, we have L.H.S. = R.H.S.
Q. 6. 4 x + 7 = 15
Sol. 4 x + 7 = 15
 4 x + 7 – 7 = 15 – 7
(Subtracting 7 from both sides)
4 x = 8
4
4
8
4
x
(Dividing both sides by 4)
x = 2
 x = 2 is the solution of the given
equation.
Check : Substituting x = 2 in the given
equation, we get
 L.H.S. = 4 × 2 + 7 = 8 + 7 = 15 and R.H.S. = 15
 When x = 2, we have L.H.S. = R.H.S.
Q. 7.
x
5
12
Sol.
x
5
12
x
5
5 12 5
(Multiplying both sides by 5)
x = 60
 x = 60 is the solution of the given equation.
Check : Substituting x = 60 in the given
equation, we get
L.H.S. 
60
5
12
 and R.H.S. = 12
 When x = 60, we have
  L.H.S. = R.H.S.
Q. 8.
3
5
15
x
Sol.
3
5
15
x
3
5
5
3
15
5
3
x
(Multiplying both sides by 
5
3
)
3
5
5
3
5 5 x
    x = 25
 x = 25 is a solution of the given
equation.
Check : Substituting x = 25 in the given
equation, we get
L.H.S. 
3
5
25
= 3 × 5 = 15
and R.H.S. = 15
  When x = 25, we have L.H.S. = R.H.S.
Q. 9. 5 x – 3 = x + 17
Sol. 5 x – 3 = x + 17
   5 x – x = 17 + 3
[Transposing + x to L.H.S. – 3 to R.H.S.]
  4 x = 20 
4
4
20
4
x
(Dividing both sides by 4)
 x = 5
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