Author: S.LAL
Created: 14 Aug, 2010; Last Modified: 29 Apr, 2018
Algebra - 03
Linear Equations in One Variable
A linear equation is an equation which involves only linear polynomialsA linear polynomial is a polynomial of degree 1, where a polynomial is an algebraic expression of two or more terms.. Equations like
7
x
+
2
=
17
and
2
x
+
5
y
+
3
z
=
25
are linear equations.
An equation involving linear polynomials of only a single variable is a linear equation in one variable or a simple equation. Some examples are:
(i)
7
x
+
2
=
17
|
(ii)
5
2
z+12=
1
2
z
|
(iii)
3
x
+
2
5
=
5
2
|
(iv)
y
−
3
4
y
=
3
y
−
16
|
Linear equation in one variable has the form:
a
x
+
b
=
0
{
 
a
,
b
∈
R
;
 
a
 
≠
 
0
 
}
.......(1)
Steps for solving simple equations
The general steps for solving any simple equation with one unknown quantity are:
- First, if necessary, clear all fractions.
- Transpose all the terms containing the unknown quantity to one side of the equation, and the known quantities to the other.
- Collect the terms on both sides.
- Divide both side by the coefficient of the unknown quantity to get the solution.
Solve the equations (i)
7
x
=
12
+
4
x
(ii)
2
3
x
+
2
=
7
4
(i)
7
x
=
12
+
4
x
⇒
7
x
−
4
x
=
12
(transposing)
⇒
3
x
=
12
(collecting terms)
⇒
x
=
12
3
(transferring numerator factor 3 to other side)
⇒
x
=
4
(ii)
2
3
x
+
2
=
7
4
⇒
2
x
+
6
3
=
7
4
Transferring denominators 3
and 4
to the other side as numerators,
⇒
4
(
2
x
+
6
)
=
3
×
7
(fractions cleared!!)
⇒
8
x
+
24
=
21
⇒
8
x
=
−
3
⇒
x
=
−
3
8
Solve the equations (i)
3
x
+
2
=
7
x
−
2
2
(ii)
1
2
y
+
2
7
y
=
2
y
+
34
(i)
3
x
+
2
=
7
x
−
2
2
⇒
3
x
−
7
x
=
−
2
2
−
2
(transposing)
⇒
−
4
x
=
−
3
2
(collecting terms)
⇒
4
x
=
3
2
(multiplying both sides with -1)
⇒
x
=
3
2
4
(ii)
1
2
y
+
2
7
y
=
2
y
+
34
⇒
7
y
+
4
y
14
=
2
y
+
34
⇒
11
y
=
14
(
2
y
+
34
)
(clearing fractions)
⇒
11
y
−
28
y
=
(
14
×
34
)
⇒
−
y
=
(
14
×
34
)
17
⇒
y
=
−
28
One third of a number increased by 4
is equal to 7
. Find the number.
Let x be the number. Then,
1
3
x
+
4
=
7
⇒
1
3
x
=
7
−
4
=
3
⇒
x
=
3
×
3
=
9
Thus, x =
9
.
The sum of two numbers is 24
and their difference is 10
. Find the numbers.
Let x be the first number. Then the second number is (24 -
x)
. So,
x
−
(
24
−
x
)
=
10
⇒
x
−
24
+
x
=
10
⇒
2
x
=
34
⇒
x
=
17
Thus, one number is 17
, while the other is (24 -
17)
or 7
.
Note that had you begun by subtracting the first number from the second, i.e.
(
24
−
x
)
−
x
=
10
, you would have got the number 7
first, from which you get 17
by subtracting from 24
.
A number consists of two digits whose sum is 8
. If 18
is added to the number, its digits are interchanged. Find the number.
[(Aggarwal & Aggarwal, Mathematics Class IX, p. 50, Ex. 1)]
Let x be ten's place digit of the number. Then, (8 -
x)
is the unit's place digit. (Note that even if you do the reverse with x in the unit's place, you will arrive at the same answer).
Since any number, like 27
, is actually
(
10
×
2
)
+
7
in expanded form, therefore the unknown number can be expressed as:
[
(
10
×
x
)
+
(
8
−
x
)
]
or
9
x
+
8
On adding 18
to the number, the digits are interchanged. This means that the new number has (8 -
x)
in the ten's place and x in the unit's place, wherein the new number can be expressed as:
[
10
(
8
−
x
)
+
x
]
or
80
−
9
x
So we have the equation:
(
9
x
+
8
)
+
18
=
80
−
9
x
⇒
9
x
+
26
=
80
−
9
x
⇒
18
x
=
54
⇒
x
=
3
Thus, with x =
3
in the ten's place and (8 -
3)
in the unit's place, the number is 35
.
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List of References
Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Bibliography
Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.