Linear Equations in One Variable
An equation contains a sign of equality, constants
and variables. If the equation has one variable and its highest power is also
one, then the equation is called a linear equation.
x = 41
A linear
equation is an equation that describes a straight line on a graph. You can
remember this by the "line" part of the name linear equation.
Standard Form of Linear Equations
A linear equation looks like this: ax + by = c
Where a, b, and
c are coefficients (numbers) while x and y are variables.
This is an
equation in two variables.
The equation in
one variable looks like this: ax + b = c
Solving Linear Equations
An equation can also be compared
with a weighing balance, the sides of the equation are the pans and the sign of
equality tells that both the pans are in balance. We represent each side of the
pan by L.H.S. and R.H.S.
Left Hand Side and Right Hand
Side are connected by the symbol of equality indicating L.H.S. = R.H.S.
On substituting a number in
place of a variable, if both the sides become equal, the number is said to be
the solution
or
root
of
the equation.
We use 3 methods to solve a
linear equation.
1.
Trial and error method
2.
Balancing method
3.
Transposition method
Trial and Error Method
In this method, we guess a number and check if it satisfies the given
equation or not. If it satisfies the given equation, then it is the solution of
the given equation otherwise we try with another number.
Example 1: Determine
whether x
= 3 is a solution of each equation.
a. x
+ 12 = 15 b. 30 – x
= 25
Solution: Substitute 3
for x
in each equation
a.. x +
12 = 15
3 + 12 = 15
15 = 15
Thus, x =
3 is a solution of the equation x + 12 = 15.
b. 30 – x
= 25
30 – 3 = 27
27 ≠ 25
Thus, x =
3 is not a solution of the equation 30 – x = 25.
Example 2: Solve each of the following equations by trial and error
method.
a. x + 4 = 10 b. 3x – 5 = 7 – x
Solution:
a. Let us think of a number
which when added to 4 gives 10. Definitely it will be 6 i.e., 6 + 4 = 10.
Hence, x = 6.
b. Let us take x = 1, then
L.H.S. = 3x – 5 = 3 × 1 – 5
= –2 and R.H.S. = 7 – x = 7 – 1 = 6
Here L.H.S. ≠ R.H.S.
Thus, x = 1 is not the
solution of the given equation.
Now, let x = 2, then
L.H.S. = 3x – 5 = 3 × 2 – 5
= 1 and R.H.S. = 7 – x = 7 – 2 = 5
Again, L.H.S. ≠ R.H.S.
Thus, x = 2 is not the
solution of the given equation.
For x = 3,
L.H.S. = 3x – 5 = 3 × 3 – 5
= 4 and R.H.S. = 7 – x = 7 – 3 = 4
Now, L.H.S. = 4 = R.H.S.
Hence, x = 3 is the
solution of the given equation.
Balancing Method
A linear equation is the same as a physical balance in its equilibrium
position. The two sides of an equation represent the two pans of a balance and
sign ‘=’ represents the beam of the balance. To solve the linear equation, we
use the principle of physical balance.
1. We can add the same number on both sides of a linear equation.
2. We can subtract the same number from both sides of a linear equation.
3. We can multiply by the same number on both sides of a linear equation.
4. We can divide by the same number on both sides of a linear equation.
Example 1: Solve: 2x – 5 = 15
Solution: 2x – 5 = 15
2x
– 5 + 5 = 15 + 5 (Adding 5 on both
sides)
2x
= 20
x
=
20/2 (Dividing
by 2 on both sides)
x
= 10
Hence, x = 10 is the
solution of the given equation.
Example 2: Solve: 5x + 3 = 23
Solution: 5x + 3 = 23
5x
+ 3 – 3 = 23 –
3 (Subtracting
3 from both sides)
5x
= 20
x
= 20/5 (Dividing by 5 on
both sides)
x
= 4
Hence, x = 4 is the
solution of the given equation.
Transposition Method
Transposition means transferring
one term to the other side. When a term is transposed to the other side, its
sign is changed.
Example 1:
Solve
3(a
–
1) = 6.
Solution: Given, 3(a – 1) = 6
3a – 3 = 6
3a = 6 + 3 (Transposing 3 to R.H.S.)
3a = 9
3a/3 = 9/3 (Dividing both sides by 3)
a = 3
Hence, a = 3 is the
solution of the given equation.
Example 2:
Solve
9.25 + 3x
=
10.75.
Solution: Given, 9.25 + 3x = 10.75
3x = 10.75 – 9.25 (Transposing 9.25 to R.H.S.)
3x = 1.50
x = 1.50/3 (Dividing both sides by 3)
x = 0.5
Hence, x = 0.5 is the
solution of the given equation.
Word Problems on Linear Equations in One Variable
Example 1:
Find
the two numbers whose sum is 25 and one of them exceeds the other by 5.
Solution: Let the smaller number
be x.
Then, the other number = x + 5
According to the question, x + x + 5 = 25
2x + 5 = 25
2x = 25 – 5 (Transposing 5 to R.H.S.)
2x = 20
x = 10 (Dividing both sides by 2)
The other number = x + 5 = 15
Hence, the numbers are 10 and 15.
Example 2: Shreya had some
ribbon. She used 96 cm of it to wrap a present. She has 208 cm of ribbon left.
How many centimeters of ribbon did she have at first?
Solution: Let x cm
represent the length of ribbon Shreya had at first.
The
required equation will be
x –
96 = 208
x –
96 + 96 = 208 + 96
x =
304
She had 304 cm of ribbon at first.
Example 3:
The
sum of three consecutive odd numbers is 129. Find the numbers.
Solution: Let the
required numbers be x, x + 2 and x + 4.
According to the problem, x + (x + 2) + (x + 4) = 129
3x + 6 = 129
3x = 129– 6 (Transposing 6 to R.H.S.)
3x = 123
3x/3
=
123/3 (Dividing both sides
by 3)
Hence, the required numbers are 41, 53 and 45.
Example 4:
The
age of a father is 30 years more than the age of his son. If the sum of their ages
is 60 years, find the ages of the father and his son.
Solution: Let the age of
the son be x
years.
The age of his father = (x + 30) years
According to the problem, x + x + 30 = 60
2x = 60 – 30 (Transposing 30 to R.H.S.)
2x = 30
x = 15 (Dividing both sides by 2)
Age of the father = (x + 30) years =
(15 + 30) years = 45 years
Hence, the age of the son is 15 years and the age of his
father is 45 years.