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Linear Equations in One Variable - Class 8 : Notes


Algebraic Expressions:
(i) Any expression containing constants, variables, and the operations like addition, subtraction, etc. is called as an algebraic expression.
(ii) Some examples of algebraic expressions are 5x, 2x – 3, x2 + 1, etc.

Equation:
(i) Any mathematical expression equating one algebraic expression to another is called as an equation.
(ii) Some examples of equations are 5x = 25, 2x – 3 = 9, x2 + 1 = 0, etc.

Linear Equation:
(i) An equation of the form ax + b = 0 where a, b are real numbers such that ‘a’ should not be equal to zero is called a linear equation.
(ii) Remember, the highest power of the variable in these expressions is 1.
(iii) Examples of some linear equations are 2x, 2x+7, 16 – 7y, etc.
(iv) Examples of some non-linear equations are x2 + 1, y + y2, etc. In these examples, the power of variable is greater than 1, thus they are non-linear equations.

Some key points:
(i) For any linear equation, there will be a presence of an equal to sign in the equation.
The quantity on left of the equality sign is called the Left Hand Side (LHS) of the equation and that on the right side is called the Right Hand Side (RHS) of equation.
For example, consider a linear equation x – 3 = 5. Here, (x-3) is the LHS of the equation and ‘5’ is the RHS of the equation.
(ii) The values of the expression on the LHS and RHS are equal and become true only for certain values of the variable. These certain values are called the solutions of the equation.
For example, for the equation x – 3 =5, the equation will be true for x = 8 i.e. for x = 8 LHS will be equal to RHS.


Solving Linear Equations:
There are two methods by which the linear equations can be solved.

(1) Balancing Method:
In this method, both the sides of equation are balanced. Let us understand it by an example:

Example:
Solve 2x - 10 = 2.
Solution: To balance both the sides, firstly we will add 10 on both the sides of the equation.
 
2x – 10 + 10 = 2 +10,
On solving, we get
2x = 12
Further, to balance the equation we will divide both the sides by 2
2x / 2 = 12/2
On solving, we get
x = 6.
Thus, x = 6 is the required solution.
Verification: Substitute the value of solution obtained in the original equation. And if after substitution, the LHS becomes equal to RHS then the solution obtained is correct.
As for above example, we obtained x = 6.
Now, let us substitute this value into the original equation 2x – 10 = 2. Then,
LHS = 2(6) – 10 = 12 – 10 = 2
RHS = 2
Here, it is observed that LHS = RHS, thus the solution obtained is correct.


(2) Transposing Method:
In this method, constants or variables are transposed from one side of the equation to other until the solution is obtained. Let us understand it by an example:

Example: Solve 2x – 5 = 5
Solution: Firstly, we will transpose the integer -5 from LHS to RHS, thus we will get
2x = 5 + 5
On solving, we get,
2x = 10
Now, we will transpose the 2 from LHS to RHS, thus we will get
x = 10 / 2
On solving, we get
x = 5.
Thus, x = 5 is the required solution.
Verification: Again, we will the substitute the obtained solution into the original equation.
Thus, LHS = 2 (5) – 5 = 10 - 5 = 5
RHS = 5
Since, it is observed that LHS = RHS, the solution obtained is correct.

 

Some Examples:
Example 1: Solve a + 5 = 15.
SolutionTransposing 5 to RHS, thus we will get
a = 15 – 5
a = 10

Example 2: Solve x/ 2 + 2 = 9/4.
SolutionLet us transpose 2 on RHS, thus we will get
x/2 = 9/4 – 2
x/2 = 1/4
Now, transposing ½ on RHS, we get-
x = 2/4 i.e. x = 1/2


Applications of linear equations:
Usually, one will encounter many practical situations where he needs to apply the methods to solve the linear equations. Let us consider an example to understand it.

Example 1: Sum of two numbers is 70. One of the numbers is 10 more than the other number. What are the numbers?
Solution: Let us assume the two numbers to be x and y.
Now, it’s given that sum of these two numbers is 70, thus, we can write
x + y = 70
Further, it is given that one of the numbers is 10 more than the other. So, let y be 10 more than x, thus, we can write-
y = x +10
Substituting value of y, we get,
x + x + 10 = 70
2x + 10 = 70
Transposing 10 on RHS,
2x = 70 – 10
2x = 60
Transposing 2 on RHS,
2x = 60
x = 30.
Now, the other number y will be x + 10 = 30 + 10 = 40 i.e. y = 40
Hence, the two desired numbers are 30 and 40.

Example 2:  The sum of three consecutive multiples is 60. What are these integers?
SolutionLet the three consecutive integers be a, a+1 and a+2.
Given, a + a + 1 + a + 2 = 60
3a + 3 = 60
Transposing 3 on LHS, we get,
3a = 60 – 3
3a = 57
On solving, a = 57/3, we get
a = 19
a + 1 = 20 and a + 2 = 21
Thus, the three consecutive integers are 19, 20 and 21.

Example 3: The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.
SolutionLet the number of winners be z. Thus, the number of participants not winning will be 63 – z.
Given, amount of winners is (100 x z) Rs i.e. 100z Rs.
Amount given to participants not winning  = 25 x (63 – z) Rs = (1575 – 25z) Rs.
Given, 100z +  1575 – 25z = 3000
Now, transposing 1575 on RHS, we get
100z – 25z = 3000 – 1575
75z = 1425
Now, dividing both the sides by 75, we get
75z / 75 = 1425 / 75
z = 19
Hence, the numbers of winners are 19.

Example 4: Rahul’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
SolutionLet the age of Rahul’s son be x. Hence, his age will be 3x.
Given, 10 years ago Rahul’s age was five times his son’s age. So, we can write
(3x – 10) = 5 (x – 10)
3x – 10 = 5x – 50
Transposing 5x to LHS and 10 to RHS, we get
3x – 5x = -50 + 10
-2x = -40
Dividing both the sides by 2, we get
So, x = 20
3x = 60.
Thus, present age of Rahul is 60 and his son’s age is 20.


1. Finding solution of Linear Equations having Variable on both the sides:
To find solution of such equations, bring all quantities with variables on side of the equation and the left out quantities on the other side. Now, solve the equation to obtain the solution.
Let us take an example.

Example 1: Solve 2x – 5 = x + 3.
Solution: Firstly, we will transpose x from RHS to LHS
2x – 5 – x = 3
Now, we will transpose the integer -5 from LHS to RHS
2x – x = 3 + 5
On solving,
x = 8

Example 2: Solve 2x + 5/3 = 26/3 – x and verify the result.
SolutionTransposing x to LHS and 5/3 to RHS, we get
2x + x = 26/3 – 5/3
3x = 21/3 i.e. 3x = 7
Diving both the sides by 3, we get
3x/3 = 7/3 i.e. x = 7/3.

Verification: Substituting values of x in LHS and RHS, we get
LHS = 2x + 5/3 = 2 (7/3) + 5/3 = 14/3 + 5/3 = 19/3
RHS = 26/3 – x = 26/3 – 7/3 = 19/3
Here, LHS = RHS. Hence, result obtained is correct.


2. Reducing Complex Linear Equations to Simpler Form:
Let us understand it by an example

Example 1: x + 7 – 8x/3 = 17/6 – 5x/2
SolutionFirstly, we will transpose 5x/2 on LHS
x + 7 – 8x/3 + 5x/2 = 17/6
Now, let us transpose 7 from LHS to RHS
x – 8x/3 + 5x/2 = 17/6 – 7
On solving both the sides, we get
5x/6 = -25/6
Thus, x= -5 is the required solution.

Example 2: Solve (7x + 4)/ (x + 2) = -4/3.
SolutionMultiplying both the sides by 3(x+2), we get
3(7x +  4) = -4(x + 2)
On solving, we get
21x + 12 = -4x -8
Transposing 4x on LHS and 12 on RHS, we get
21x + 4x = -8 -12
25x = -20
Dividing both sides by 25, we get
x = - 4/5.

Example 3: The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2 . Find the rational number.
SolutionSuppose, the numerator of rational number is x. hence, its denomiantor will be x + 8.
Then the rational number will be x/ (x+8).
Given, .if the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2, thus we get
(x + 17) / (x + 8 – 1) = 3 / 2
(x + 17) / (x + 7) = 3 / 2
Multiplying both the sides by 2(x + 7), we get
2(x + 17) = 3(x + 7)
2x + 34 = 3x + 21
Transposing 3x on LHS and 34 on RHS, we get
2x – 3x = 21 – 34
-x = -13 i.e. x = 13
x + 8 = 13 + 8 = 21.
Thus, the rational number will be 13/21.



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