Introduction to Linear Equations in One Variable Class 8 Notes | EduRev

Mathematics (Maths) Class 8

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Class 8 : Introduction to Linear Equations in One Variable Class 8 Notes | EduRev

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

We are well aware of the term “Equations”. An equation is an expression that equates with another. For example,

  • 3x = 9
  • 2x+4 = 10
  • 3xy +7 = 40
  • abc+b+c = 15

The expressions in the example given above signify expression of an equation in terms of variables. The expressions with only one variable are called Linear Equations in one variable. These expressions have only one variable. Variables are the numbers mentioned in terms of any alphabets, in the examples given we see variables like a,b,c,x,y.
Introduction to Linear Equations in One Variable Class 8 Notes | EduRev
Algebra Equations:
Expressions that involve variables in equality are called algebra equations. These expressions are formed by two equations involving expressions on left and right with an equality sign in between the two. The expression on the left is called Left Hand Side (L.H.S) and the expression on the right is the Right Hand Side expression (R.H.S).

Generally, in an equation, the values of the L.H.S expression is equal to that on the R.H.S and these values are called their solutions. To obtain the solutions of an equation we use the various arithmetical operations on the two sides so that the balance between the two equations are not disturbed.

Solving Various Types of Linear Equations in One Variable:
Solving Linear equations in one variable involves a few techniques of balancing the two expressions on both sides of the equal sign.
Technique 1
For understanding this technique lets consider the example given below:
4x - 5 = 15
Adding 5 to both sides, we get
4x-5+5 = 15+5
4x=20
Now, divide both sides with 4,
4x/4 = 20/4
x=5

Technique 2
The following example will help us understand the technique in a better way: 3z+8 = 10
Changing the position of 8 to R.H.S,
3z = 10-8
When we move a number from one side to another, the sign before the number inverts to the opposite sign. – sign becomes +; × sign becomes ÷ and vice-versa. 
Now, 3z=2
⇒ z= 2/3
Let’s check our answer by equating the two sides and replacing the solution with the variable.
3z+8 = 10
3(2/3) + 8 = 10
2+8=10
10=10
L.H.S = R.H.S
Hence the value of z  is accurate.


Solving Linear Equations having Variables on Both Sides:
In the topic above we learned about the equations with numbers on one side. Here we shall learn equations with expressions on both sides. For solving such equations, the techniques used above shall be followed, but with care and accuracy of operations. The following examples will make our understanding of the same clear:
2a-5 = a+4
2a= a+4+5
or, 2a =a+9
Now, we subtract a on both sides;
2a – a = a+9-a
a = 9
The point to be noted here in the example above is that we have subtracted a variable “a” from both sides. In an algebraic equation, a variable is also considered a number as in a way it represents a number. Let’s take another example:
3y + 5/2 = 3y/2-14
Let’s multiply both sides with 2 ( as it is the denominator on both sides)
2× ( 3y + 5/2 ) = 2× ( 3y/2-14)
and (2×3y) +2×(5/2 ) = (2×3y/2)-(2×14)
6y +5 = 3y- 28
We change the position of 3y from RHS to LHS,
6y-3y+5 = - 28
3y + 5 =- 28
3y = -28 - 5
y = -33/3
y=-11

Reducing a Linear Equation to Simpler Forms:
Linear equations may not always be simple! We may come upfront with equations that are a combination of many operations. In such a case we apply the basic rule of B.O.D.M.A.S. The following example will make things clear:
25a - 4(4a - 3) =4(3a - 2) + 7/2
The first step here involves opening of all the brackets. Let’s take the expression on LHS:
25a - 4(4a - 3)
= 25a - 16a + 12
=9a + 12
Now let’s take RHS,
4(3a - 2)+7/2
=12a -8 +7/2
=12a - 16/2+7/2

=12a - 9/2
Now the equation is; which includes the simplified expression of LHS and RHS
9a +12 =12a - 9/2
12 = 12a - 9a - 9/2
12 = 3a - 9/2
12+9/2 = 3a
24/2 +9/2 = 3a
33/2  = 3a
33/2 × 1/3 = a
a =11/2
Now, let’s check our solution by replacing the solution with the variable:
25a - 4(4a -3) =4(3a-2) + 7/2
LHS = 25 × 11/2 – 4 (4 ×11/2 -3)
=275/2 – 4 (22-3)
=275/2 – 76 = 275/2 -152/2 =123/2
RHS = 4(3a-2)+7/2
=4 (3×11/2 – 2) +7/2 =4(33/2-2) +7/2
=4(29/2)+ 7/2= 58+7/2= 116/2+7/2
= 123/2
The answer on both sides is equal, hence the solution is proved. You must have noted in the above equations that, while solving a linear equation in one variable, we take into consideration the expression on both the sides one by one. This not only makes the concept clear but also helps us to reach to solutions faster.


Application of Linear Equations:
Linear equations in one variable can be used to calculate the value of an integer when we are given two or more integers with one of them being unknown. Using the expression in an equation form, we can easily find the value of an unknown integer.

Linear equation helps us deduce integers in many word problems of arithmetic. Not only these expressions make the equations simple to understand but also help in reaching to conclusions easily and fast.


Solved Example for You

Q. In the last three months Mr.Sharma lost 5 and a half kg, gained 2 and one-fourth kg and then lost 3 and three fourth kg weight. If he now weighs 95 kg, then how much did Mr.Sharma weighs in beginning?

A. 100 kg
B. 102 kg
C. 106.5 kg
D. 104 kg

Ans: B.
Let x be the Mr.Sharma’s weight in beginning.
According to given in instructions,
x - 5.5 + 2.25 - 3.75 = 95
x - 7 = 95
Transposing 7 to R.H.S, we get
x = 102 kg

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