Linear equations in one variable for class 8

“Tired about the chapter linear equations in one variable “? Here you will learn the complete concept which will help you to solve the problems.In this article first, you will learn concept and then solution of some problems based on that concept. At the there are some practicing and challenging problems which you should must try. Here are topic which we will learn in this article

  • What is the equation?
  • What is the linear equation in one variable?
  • What is the transposition method?
  • How to solve a linear equations using transposition?
  • word problems
  • practice questions

What is the equation?

An equation is a statement in which one quantity is equal to other quantity .this equality is indicated by symbol equal (=) which was discovered with mathematician name Robert recorder.

e.g          3x+9=0 is equation because left-hand side (LHS) is equal to right-hand side(RHS)

Note:  There are several types of equations like the linear equation, quadratic equation,, cubic equation……etc but in this article, you will only learn about linear equations in one variable other you will learn in higher classes.

What is a linear equation?

An equation whose highest power of the variable is one is called a linear equation.

e.g      3x+9=0 ,3x+4y=5 are linear equations

but 3x²+9 is not linear because the highest power of the variable is 2.

What is the linear equation in one variable?

The linear equation which involves only one type of variable is called a linear equation in one variable.

e.g            3x+9=0 is a linear equation in one variable because it has only a variable x.

but         3x+4y=5   is not a linear equation in one variable because it has two variables x and y.

What is the transposition method?

Transposition means transferring position .we use it to find the solution.

when we transfer any term from one side to another side its sign changes. Here is its rule

        (-)            is changed to (+)

                  (+)  is changed to  (-)

(×)   is changed to (÷)

(÷) is changed to (×)

Before learning the procedure to solve the linear equation, let’s know about the solution of the equation.

What is the solution of the equation

The value of the variable which satisfies the equation is called solution or root of the equation

e.g               3x+9=0

when we substitute x=-3 ,then LHS=RHS

∴      -3 is a solution of the equation

How to solve linear equations in one variable using transposition?

We can easily solve any linear equation in one variable using the transposition method

Let us understand with the example

Q.1 solve the equation 3x+7=16 and verify the answer.

solution: we have,                3x+7=16

3x=16-7 (transposing 7 to RHS)

3x=9

x=3

3 is the solution of the equation.

verification:  If 3 is the solution of the equation, so it must satisfy the equation

put x=3 in equation 3x+7=16

LHS=3x+7=3(3)+7=16RHS

since     LHS=RHS

∴ 3 is a solution of the equation

Q.2 solve the equation for m

4(3m+1)-2(2m+3)=3(m+4)+2m+7

solution:  we have,

4(3m+1)-2(2m+3)=3(m+4)+2m+7

                                 12m+4-4m-6=3m+12+2m+7         (simplifying both side)

8 m-2   =   5m+19

8m-5m=19+2           (Transposing 5m to LHS and -2 to RHS)

3m=21

m=7

so 7 is a solution of the equation

Word problem based on linear equations in one variable

Q.1 If 3 is subtracted from a number and the result is multiplied by 5 , we obtain 15 .what is the number?

solution.we have three-part of question

1.3 is subtracted from a number

2.result is multiplied by 5

3.obtain 15

Let the required number is x

According to the first part,3 is subtracted from a number

we obtain the result            x-3

part.2 result is multiplied by 5

∴          5(x-3)

part.3 the result obtain is 15

5(x-3)=15

5x-15=15

5x=30

x=6

6 is the required number

Q.2 The perimeter of a rectangular park is 200 m. Its length is 10 m more than twice its breadth .what is the length and breadth of the park?

solution: we have only part of this question

1. Its length is 10m more than twice its breadth

2. The perimeter of a rectangular park is 2oom

Let the breadth of the rectangle be x m

Part.1 The length of rectangle 10 m more than twice of the breadth

length =2x+10      ( twice of breadth =2x)

 

part.2 the perimeter of rectangle is 200m

we know,

perimeter of rectangle =2(l+b)

200=2(2x+10+x)   [perimeter=200,length=2x+10 and

200=2(3x+10)                           breadth=x]

3x+10=100

3x=90

x=30

breadth of rectangle=30m

length of rectangle=2x+10=2(30)+10=60+10=70m

Q.3 The sum of three consecutive multiples of 6 is 666.find these multiples

solution. Let three consecutive multiples of 6 be x,x+6, and x+12

sum of these multiples =666

according to the question,

x+(x+6)+(x+12)=666

3x+18=666

3x=648

x=216

Hence,three consecutive multiples of 6 are 216

x+6=216+6=222

x+12=216+12=228

Q.4 Rohit is now one-third of his father’s age. After twelve years, the age of Rohit’s father will be twice the age of Rohit. Find their present ages.

solution:  Let Rohit’s present age be x years

present age of his father =3x years

Age of Rohit’s after 12 years =(x+12) years

Age of Rohit’s father after 12 years =(3x+12)years

∴ By the given condition

3x+12=2(x+12)

3x+12=2x+24

x=12

present age of Rohit=12 years

present age of Rohit’s father=36 years

Some practice problems on linear equation in one variable

Here are some practice problems which you should try to solve. If you can solve then congratulation to you otherwise comment  your problem with any question , I will reply to you.

In the end, there are two challenging problems which you should also try.

Q.1 The length of a rectangle is four times its breadth and the perimeter of the rectangle is 90m.Find the dimension of the rectangle.

Q.2 Present ages of veena and simi are in the ratio 2:5 . Four years from now the ratio of their ages will be 1:2 . Find their present ages.

Q.3 15 years from now Renu will be four times her present age. What is Renu’s present age?

Q.4 Meena has 600 Rs. in the form of notes of 10 rs and 20 rs denomination. The number of 10 rs notes is three times the number of 20 rs . Find the number of notes of each denomination.

Q.5 The sum of the two numbers is 3000. If 8% of one number is equal to 12% of the other, Find the numbers.

ANSWER

1.36m and 9m

2.8 years and 20 years

3.5 years

4.36 notes of 10 rs,12 notes of 20 rs

5.1800 and 1200

Challenging problems

Q.1The age of the father is twice the sum of the ages of his two children . After 20 years , his age will be equal to the sum of the ages of his children. What is father’s age?

Q.2 In the examination , one mark is awarded for every correct answer while the 1/2 mark is deducted for every wrong answer. A candidate answered 80 questions and got 60 marks . How many questions did he answer correctly?

 

 

 

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