# Linear Equations in One Variable Notes | Study Mathematics (Maths) Class 8 - Class 8

## Class 8: Linear Equations in One Variable Notes | Study Mathematics (Maths) Class 8 - Class 8

The document Linear Equations in One Variable Notes | Study Mathematics (Maths) Class 8 - Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8

Linear equation is an algebraic equation that is a representation of the straight line. Linear equations are composed of variables and constants. These equations are of first-order, that is, the highest power of any of the involved variables i.e. 1. It can also be considered as a polynomial of degree 1. Linear equations containing only one variable are called homogeneous equations. The corresponding variable is called the homogeneous variables. For instance,

• x + 2y = 3 is a linear equation in two variables.
• x + y + z = 8 is in three variables.
• x + y2 = 1 is not a linear equation because the highest power of y is 2.

Linear Equation

Standard Form of Linear Equation in One Variable
A linear equation in one variable can be expressed in the form of ax + b = 0, where x is the variable and a and b are the constants involved. These constants (a and b) should be non-zero real numbers. These types of equations have only one possible solution for the value of the variable.

Steps for Solving Linear Equations in One Variable

The following steps are performed to solve linear equations in one variable:
Step 1: In case the integers a and b are fractional numbers, LCM has to be taken to clear them.
Step 2: The constants are taken to the right side of the equation.

Solving Equations which have Linear Expressions on One Side and Numbers on the Other Side

Linear equation is an algebraic equation that is a representation of the straight line. Linear equations are composed of variables and constants. These equations are of first-order, that is, the highest power of any of the involved variables i.e. 1. It can also be considered as a polynomial of degree 1. Linear equations containing only one variable are called homogeneous equations. The corresponding variable is called the homogeneous variable.

Solving a system of linear equations in one variable
The linear equations in one variable are represented in the form ax+b = 0 where x is a variable, a is a coefficient and b is a constant. These equations can be solved by the following steps:
Step 1: In case the integers a and b are fractional numbers, LCM has to be taken to clear them.
Step 2: The constants are taken to the right side of the equation.
Step 3: All the terms involving the variable are isolated to the left-hand side of the equation, to evaluate the value of the variable.
Step 4: The solution is verified.

Sample Problems on Linear Equations

The following examples illustrate the complete procedure of solving algebraic expressions composed of variables on one side (L.H.S) and constants on other (R.H.S).

Examples 1: For equation 5x – 20 = 100, find the value of x?
Solution:
Transposing 20 to RHS, we have,
5x = 120
Dividing both side by 5,
5/5 x = 120/5
x = 24, which is the final solution

Example 2: For equation 4/3z + 1/9 = -1, find the value of z?
Solution:
Transposing 1/9 to RHS,
4/3 z = -1 -1/9
⇒ 4/3 z = -8/9
Multiplying both sides by 3/4,
⇒ 3/4 x 4/3 z = -8/9 x 3/4
z = -2/3, which is final solution

Example 3: For equation 9/10 + y = 3/2, find the value of y?
Solution:
Transposing 9/10 to RHS, we get,
y = 3/2 – 9/10
Taking LCM of RHS
y = (15-9)/10
y = 6/10
Simplifying, we have
y = 3/5 , which is the final solution.

Note: The solution for the variable may be integers or rational numbers.

Example 4: Ben’s age is 4 times the age of William. 10 years earlier, Ben’s age was 14. What is his William’s current age?
Solution: Let William’s current age be w.
Now , Ben’s current age is 4 times the age of William, which is equivalent to 4w.
Ben’s age 10 years earlier = Ben’s current age – 10
= 4w -10
Now, according to the question,
4w – 10 = 14
4w = 24
w = 6 years
Therefore, William’s current age is 6 years.

Example 5: Aman bought 5 chocolates for Rs 35. What is the cost of each chocolate?
Solution: Let the cost of each chocolate be r Rs.
Now, according to the question,
5x = 35
x= Rs 7
Therefore, the cost of each chocolate is Rs 7.

Example 6: Sita had some tiles and Gita had 1/8 of the tiles of Sita. The difference in their tiles is 4. What is the number of tiles with Gita?
Solution: Let the number of tiles with Sita be x.
Now, Gita has 1/8 x tiles.
According to the question ,
SIta’s tiles – Gita’s tiles = 4
Therefore,
x – 1/8x = 4
7/8x = 4
x = 32/7 tiles
Now, Gita has 1/8 * 32/7 tiles = 4/7 tiles.

Solving Linear Equations with Variable on Both Sides

Equations contain two types of quantities, one variable the other is number. But we can consider both equivalents for the sake of reverse operations. This is one of the important things that a student must keep in mind for solving various problems. Also, a good understanding of reverse operations would lead to a better level of problem solving.
In some questions, the variable will appear on both sides of an equation. To solve this kind of questions we need to note the following points:

• We can add a number with a variable to both sides without changing the equation or the values.
• We can subtract a number with a variable to both sides without changing the equation or the values.
• We can multiply a number with a variable to both sides without changing the equation or the values.
• We can divide a number with a variable to both sides without changing the equation or the values.

Example: Solve 14 – 2x = 5x for the value of x.
Step 1: First, we need separate variables on one side and number on other side by applying some basic operations.
Add 2x on both the sides
14 – 2x + 2x = 5x + 2x
(Similarly, we can subtract a term with a variable from both sides of the equation)
Step 2: Perform operations to convert the co-efficient of the variable to 1.
Equation: 14 = 7x
Divide 7 on both the sides
x = 2

Example: Solve 64 + 2x = 10x + 8 for the value of x
Step 1: Subtract 2x from both sides:
64 + 2x – 2x = 10x – 2x + 8
64 = 8x + 8

Step 2: Subtract 8 from both the sides:
64 – 8 = 8x + 8 – 8

56 = 8x

Step 3: Divide 8 on both the sides
x = 7
Note: In every problem of this kind it is always recommended separating the numbers and variables on either side of the equation by applying the reverse operations.

Sample Problems on Linear Equations
Example 1. Solve for x: 35x – 45 = 25
Solution:
Add 45 to both the sides
35x – 45 + 45 = 25 + 45
35x = 70
Divide 35 on both the sides
x = 2

Example 2. Solve for x: 22 – 32x = 33 + x
Solution:
Add 32x on both the sides
22 – 32x + 32x = 33 + x + 32x
22 = 33 + 33x
Subtract 33 from both the sides
22 – 33 = 33 + 33x -33
-11 = 33x
Divide 11 on both the sides
-1 = 3x
Divide 3 on both the sides
x = -1/3

Example 3. Solve for x: 23x + 4 = 104 + 3x
Solution:
Subtract 4 from both the sides
23x + 4 – 4 = 104 + 3x – 4
3x = 100 + 3x
Subtract 3x from both the sides
23x – 3x = 100 + 3x – 3x
20x = 100
Divide 20 from both the sides
x = 5

Example 4. Solve for x: 45x + 21 = 15x + 141
Solution:
Subtract 21 from both the sides
45x + 21 – 21 = 15x + 141 – 21
45x = 15x + 120
Subtract 15x from both the sides
45x – 15x = 15x + 120 – 15x
30x = 120
Divide 30 on both the sides
x = 4

Example 5. Solve for x: 28x + 33 = 108 + 3x
Solution:
Subtract 3x from both the sides
28x + 33 -3x = 108 + 3x – 3x
25x + 33 = 108
Subtract 33 from both the sides
25x + 33 – 33 = 108 – 33
25x = 75
Divide 25 on both the sides
x = 3

Example 6. Solve for x: 8x + 3x = 34 + 2 + 2x
Solution:
Simplify: 11x = 36 + 2x
Get the variable on one side:
11x – 2x = 36 + 2x – 2x
9x = 36
Solve using inverse operations:
x = 4
Check Whether: 8(4) + 3(4) = 34 + 2 + 2(4)?
Yes!

Example 7. Solve for y: 33y – 32 = 19 – 18y
Solution:
Get the variable on one side using inverse operations
33y – 32 + 18y = 19 – 18y + 18y
51y – 32 = 19
51y – 32 + 32 = 19 + 32
51y = 51
y = 1
Check: 33y – 32 = 19 – 18y?
Yes!

The document Linear Equations in One Variable Notes | Study Mathematics (Maths) Class 8 - Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8.
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