Class VI�CHAPTER 11�TOPIC: ALGEBRA�PART- 2 � (NCERT Syllabus)

PREPARED BY: MRS. KALPITA J. ZOPE

T.G.T. MATHS

JNV SATARA (MAHARASHTRA)

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Content:

• What is Equation
• Solution of an Equation
• Trial and Error method for finding the solution of an equation
• Examples
• What we have learnt
• Assignments : Multiple choice question

Practice Questions

PISA Based Question

What is an Equation :

• We all know about a balance which a vegetable

seller use in which two plates are balance with

a horizontal,

• If equal quantities are placed in both sides of the

balance only then it is called in balance position.

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• Balance position

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• If we put 4 balls in right plate and some balls

without counting in left plate, which is heavier

than right plate.

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• Let number of balls in left plate = x

now x > 4

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x balls 4 balls

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• If we remove one ball from left plate

again x – 1 > 4

• Here x – 1 is an algebraic expression

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(x -1) balls 4 balls

• If we remove 1 more ball from left plate now both plates are in equal level, so, x – 2 = 4

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• Here x – 2 = 4 is called an equation in which x – 2 is left hand side (LHS) of equation and 4 is right hand side (RHS) of equation

Both is bounded with equal sign so it is called an Equation.

Solution of an Equation:

• Let us recall the matchstick pattern of L.

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• We can prepare general rule for this

Number of matchsticks = 2 x number of L’s = 2n

where n is a variable.

• If it is given that,

Number of matchstick = 10

So, 2n = 10 is an equation

For finding solution of equation

We have to find value of n (Number of L’s) for which LHS = RHS for given equation

For n = 3, LHS = 2n = 2x3 = 6 ≠ 10 (RHS)

For n = 4, LHS = 2n = 2x4 = 8 ≠ 10 (RHS)

For n = 5, LHS = 2n = 2x5 = 10 = RHS

For n = 5 LHS = RHS, So n = 5 is a solution of equation 2n = 10

The value of variable which gives both sides of equation are equal is called its solution.

OR For a fixed value of variable which satisfy the equation is called

the solution of equation.

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Trial and Error method for finding the solution of an Equation:

• Taking different values of variable to find the solution of equation is called solution of equation by trial and error method.
• There are other methods also for finding solution of an equation which we study in upper classes.

• EXAMPLE: Solve equation 3x – 2 = 10

By Trial and Error method,

Take x = 1

LHS = 3x – 2 = 3 x 1 – 2 = 3 – 2 = 1 ≠ 10 ( RHS)

Take x = 2

LHS = 3x – 2 = 3 x 2 – 2 = 6 – 2 = 4 ≠ 10 ( RHS)

Take x = 3

LHS = 3x – 2 = 3 x 3 – 2 = 9 – 2 = 7 ≠ 10 ( RHS)

Take x = 4

LHS = 3x – 2 = 3 x 4 – 2 = 12 – 2 = 10 = RHS

Therefore, For x = 4 LHS = RHS

Therefore, x = 4 is a solution of given equation.

What we have learnt:

• Some times numbers can be represented as variables.
• A variable takes on different values, its value is not fixed.
• We can use any letter of English alphabet to show variable.

Example – x, y, z, n, l, m, p etc.

• Variable can use to express the general rules.
• Algebraic expressions having variables and numbers with mathematical operations.
• Value of an algebraic expression is change as value of variable varies.
• Equation is an equality having two sides (LHS and RHS)
• We can find solution of an equation by using trial and error method.
• Value of variable which satisfy the equation is called the solution of equation.

ASSINGNMENT:��MULTIPLE CHOICE QUESTIONS:

1. Which of these is not an algebraic expression

(a) 3x (b) 5y – 2 (c) 2 x 3 – 2 (d) 2m + 3

2. Half of a x increased by 4 can be returned as

(a) 2x + 4 (b) ½ + 4 ( c ) x + ½ (d) x/ 2 + 2

3. Twice a number decreased by 4 is written as

(a) 2x + 4 (b) 4x – 2 ( c ) 2x – 4 (d) 4x + 2

4. Sonu is x year old. His mother is 10 yr more than his age . How old is mother ?

(a) (x + 10) yrs (b) (2x + 10) yrs ( c ) ( 2x – 10 ) yrs

(d) 10x yrs

5. If a regular polygon has a n sides each measuring s units, its perimeter is expressed as

(a) (b) 4n ( c ) n x s (d) n + s

PRACTICE QUESTIONS:

1. Each side of a square is k units. Express its perimeter and area in algebraic form .

2. Ritu buys 5 copies for Maths , 2x copies for English and 2y copies for Hindi. Express the total number of copies she buys as an algebraic expressions.

3. Write algebraic expressions for the following

(a) x is divided by 12 and 23 is added to a result.

(b) Price of tea per is Rs 10 less than that of coffee(c).

4. Write statement for (a) 8v + 5 (b) 9y

5. Find the value of algebraic expression 2y + 7 for y = 1 and y = 2.

6. If x = 2 , find the value of 2x – 1

7. Find solution for following equation by trial and error method

(a) y – 8 = 6 (b) 2x + 1 = 15

8. Prepare equation for the statement “ 5 more than a number equals 11.”

PISA BASED QUESTIONS:�

Seeta and Geeta have some money, Seeta says to Geeta “ Your amount is 5 less than 2 times of mine”.

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Seeta Geeta

With the above statement answer the following:

1. Assume that Seeta’s amount as a variable.
2. Write Geeta’s amount as an algebraic expression.
3. If Seeta’s amount is Rs. 20, then find Geeta’s amount.
4. If Geeta’s amount is Rs. 15, then prepare an equation for this and find Seeta’s amount in this situation.

THANK YOU