LINEAR EQUATIONS EX 2.2 (LECTURE 3) 22.04.2020

LINEAR EQUATION CLASS 8 EX 2.2 (I)  22.04 20

Good morning students!!!!!!!!!!!!

Guidelines for the class:

1. Note down  the work in your register on a regular basis.

2. Please feel free to ask, if you have any doubt by dropping a message in the comment box.

3. Make a column on the right hand side, if you need to do any rough work

4. Write  today's date.

5. Write the chapter number and name.

6. Text in red has to be noted down as its your class work.

7. Text in green has to be read thoroughly and understood.

In the previous lectures we have already seen solving equations which have Linear Expressions on one side and Numbers on the other side.

Linear Equation in One Variable Definition: 

A linear equation in one variable is an equation which has a maximum of one variable of order 1. It is of the form ax + b = 0, where x is the variable.This equation has only one solution. 

Few examples are:

• 3 x = 1
• 22 x -1=0
• 4 x+9 = -11

Standard Form of Linear Equation in One Variable

The standard form of linear equations in one variable is represented as: 

ax + b = 0

Where,

• ‘a’ and ‘b’ are real numbers, and
• both ‘a’ and ‘b’ are not equal to zero.

Thus, is the formula of linear equation in one variable is ax + b = 0.

Solving Linear Equations in One Variable

For solving an equation having only one variable, the following steps are followed

• Step 1: Using LCM, clear the fractions if any.
• Step 2: Simplify both sides of the equation.
• Step 3: Isolate the variable.
• Step 4: Verify your answer.

Example of Solution of Linear Equation in One Variable

Let us understand the concept with the help of an example.

For solving equations with variables on both sides, the following steps are followed:

Consider the equation: 5x – 9 = -3x + 19

Step 1: Transpose all the variables on one side of the equation. By transpose, we mean to shift the variables from one side of the equation to the other side of the equation. In the method of transposition, the operation on the operand gets reversed.

In the equation 5x – 9 = -3x + 19, we transpose -3x from the left-hand side to the right-hand side of the equality, the operation gets reversed upon transposition and the equation becomes:

5x – 9 +3x = 19

⇒ 8x -9 = 19

Step 2: Similarly transpose all the constant terms on the other side of the equation as below:

8x -9 = 19

⇒ 8x = 19 + 9

⇒ 8x = 28

Step 3: Divide the equation with 8 on both sides of the equality.

8x/8 = 28/8

⇒ x = 28/8

If we substitute x = 28/8 in the equation 5x – 9 = -3x + 19, we will get 9 = 9, thereby satisfying the equality and giving us the required solution.

Today, we will see some applications of the above method in our daily life situations.

How to crack word problems:

     

● Read the linear problem carefully and note what is given in the question and what is required to find out. 

● Denote the unknown by any variable as x, y, ……. (any variable) 

● Translate the problem to the language of mathematics or mathematical statements. 

● Form the linear equation in one variable using the conditions given in the problems. 

● Solve the equation for the unknown. 

● Verify to be sure whether the answer satisfies the conditions of the problem

Linear Equations in One Variable Word Problems 

(Do it in the register)

Problem: 1 The length of the legs of an isosceles triangle is 4 meters more than its base. If the Perimeter of the triangle is 44 meters, find the lengths of the sides of the triangle.

Solution:

Let us assume the base measures ‘x’ meter. Hence each of the legs measure y = (x + 4) meters.

The Perimeter of a triangle is the sum of the three sides.

The equations are formed and solved as follows:

x + 2(x + 4) = 44

x + 2x + 8 = 44

3x + 8 = 44

3x = 44 – 8 = 36

3x = 36

x = 36/3

x = 12

The length of the base is solved as 12 meters. Hence each of the two legs measure 16 meters.

Problem:2 The sum of three consecutive multiples of 4 is 444. Find these multiples. 

Solution: 

If x is a multiple of 4, the next multiple is x + 4, next to this is x + 8. 

Their sum = 444

According to the question, 

x + (x + 4) + (x + 8) = 444 

⇒ x + x + 4 + x + 8 = 444

⇒ x + x + x + 4 + 8 = 444 

⇒ 3x + 12 = 444

⇒ 3x = 444 - 12 

⇒ x = 432/3 

⇒ x = 144

Therefore, x + 4 = 144 + 4 = 148 

Therefore, x + 8 - 144 + 8 – 152

Therefore, the three consecutive multiples of 4 are 144, 148, 152.

Problem:3 The denominator of a rational number is greater than its numerator by 3. If the numerator is increased by 7 and the denominator is decreased by 1, the new number becomes 3/2. Find the original number.
Solution:

Let the numerator of a rational number = x

Then the denominator of a rational number = x + 3

When numerator is increased by 7, then new numerator = x + 7

When denominator is decreased by 1, then new denominator = x + 3 - 1

The new number formed = 3/2

According to the question,

(x + 7)/(x + 3 - 1) = 3/2

⇒ (x + 7)/(x + 2) = 3/2

⇒ 2(x + 7) = 3(x + 2)

⇒ 2x + 14 = 3x + 6

⇒ 3x - 2x = 14 - 6

⇒ x = 8

The original number i.e., x/(x + 3) = 8/(8 + 3) = 8/11.

Now let us do Exercise 2.2  (Do all these sums in your register)

Question 2:  The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth?

Ans. Let the breadth of the pool be  m.

Then, the length of the pool = 2x+2 m

Perimeter = 

 154 = 

 

[Dividing both sides by 2]

 

 

[Subtracting 2 from both sides]

 

 25=x

[Dividing both sides by 3]

 

 m

Hence, length of the pool = 

= 50 + 2 = 52 m

And,    breadth of the pool = 25 m.

Question 4: Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Ans. Sum of two number = 95

Let the first number be  then another number be .

According to the question, 

 

 

[Subtracting 15 from both sides]

 

 

[Dividing both sides by 2]

 

Hence, the first number = 40

And another number = 40 + 15 = 55.

Question 5: Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?

Ans. Let the two numbers be  and 

According to question,  

 

 

[Dividing both sides by 2]

 

Hence, first number =  = 45 and second number =  = 27.

Question 8:

(Do it in the register)

Question 9: The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?

Ans. Let the present ages of Rahul and Haroon be  years and  years respectively.

According to condition,

 5x+7x+4+4=56

 12x+8=56

[Subtracting 8 from both sides]

 

 

[Dividing both sides by 12]

 

Hence, present age of Rahul =  = 20 years and present age of Haroon

=  = 28 years.

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Question 10: The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?

Ans. Let the number of girls be 

Then, the number of boys = 

According to the question,

 

 5x+40=7x

 

[Transposing  to L.H.S. and 40 to R.H.S.]

    

[Dividing both sides by -2]

 x=20

Hence the number of girls = 20 and number of boys = 20 + 8 = 28.

Linear Equations in One Variable Word Questions (Worksheet)A few practice questions are given below.

• Question 1: Solve ( 10x – 7) = 21
• Question 2: Find the multiples, if the sum of two consecutive multiples of 6 is 68.
• Question 3: Verify that if x = -3, is a solution of the linear equation 10x + 7 = 13 – 5x.

 HOMEWORK

Exercise 2.2 : Qns: 1,3, 6 and 7.
Please refer SDG 8  and write down in your register.

That's all for today!! See you in the next class, will continue the same exercise!!!

Good Morning and Thank you boys.