RS Aggarwal Solutions: Linear Equation in One Variable

# RS Aggarwal Solutions: Linear Equation in One Variable | Mathematics (Maths) Class 6 PDF Download

``` 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
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
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
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
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
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
```

## Mathematics (Maths) Class 6

94 videos|347 docs|54 tests

## FAQs on RS Aggarwal Solutions: Linear Equation in One Variable - Mathematics (Maths) Class 6

 1. What is a linear equation in one variable?
Ans. A linear equation in one variable is an equation that can be written in the form ax + b = 0, where 'x' is the variable, 'a' and 'b' are constants, and 'a' is not equal to 0.
 2. How do you solve a linear equation in one variable?
Ans. To solve a linear equation in one variable, follow these steps: 1. Simplify the equation by removing any parentheses or fractions. 2. Combine like terms on both sides of the equation. 3. Get rid of any constants on one side by adding or subtracting. 4. Isolate the variable on one side by performing inverse operations. 5. Solve for the variable by dividing both sides by the coefficient of the variable. 6. Check the solution by substituting it back into the original equation.
 3. Can a linear equation have more than one solution?
Ans. No, a linear equation in one variable can have at most one solution. This means that there can be only one value of the variable that satisfies the equation. If there are multiple solutions, it means that the equation is not linear or that it has been incorrectly solved.
 4. What is the importance of linear equations in one variable?
Ans. Linear equations in one variable are important in various fields of mathematics and real-life applications. They help in solving problems involving unknown quantities and relationships between variables. They are used in financial calculations, physics, engineering, and many other areas where relationships can be represented by a straight line.
 5. Can a linear equation in one variable have no solution?
Ans. Yes, a linear equation in one variable can have no solution. This happens when the equation leads to a contradiction, such as 2 = 3 or 0 = 1. In such cases, there is no value of the variable that satisfies the equation, and it is said to have no solution.

## Mathematics (Maths) Class 6

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