Linear Equations in One Variable

# Linear Equations in One Variable

## 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.

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)
x = 41
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.