WELCOME CHAPTER: 3 PAIRS OF EQUATIONS - PowerPoint Presentation

Slide1

WELCOME

Slide2

CHAPTER: 3

PAIRS OF EQUATIONS

STANDARD: 9

Slide3

Consider the problem

,

“ There are 100 beads in a box, some black and some white; 10 more black than white. How many black, how many white?”

MENTAL MATH AND ALGEBRA

Slide4

We can do it in several ways.

F

irst method:

( calculation in mind)

Imagine that we take the extra 10 black beads , then we have 90 left in the box in which black and white are equal. So 45 of each. By adding the 10 black bead kept apart, black become 55 and white remains at 45.

Slide5

Algebraic method:

Let the number of black beads as x, the number of white beads is x-10 and since there are 100 in all

,

x

+(x-10) = 100

2x-10 = 100

2x =110

x=55

So the number of black beads = 55;

Subtracting 10, we get the number of white beads = 45.

Slide6

Another way using Algebraic method

:

Let number of black beads = xAnd number of white beads = y.

Then,

x + y = 100

x - y = 10

We can solve this as

2x=(x+y)+(x-y) = 110

2y=(x+y)-(x-y) = 90

So,

x

=55 & y = 45

Slide7

Consider another problem,

“ A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of boys. How many boys and girls are there in the class?”

Slide8

Answer:

Let the number of boys = x

Then number of girls = x+10

On a particular day,

Number of boys present = x -8

We have,

2(x-8)= x+10

2x-16= x+10

x= 26

So, number of boys = 26

& number of girls = 26+4 = 30

Slide9

Consider the problem,

“ The price of 2 pens and 3 notebooks is 40 rupees and the of 2 pens and 6 notebooks is 60 rupees. What is the price of a pen and a notebook?”

TWO EQUATIONS

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Let the price of a pen = x rupees and the price of a notebook = y.

Then we have the following equations:2x+3y = 402x+5y = 60

Then we have,

2y = 20

Y = 10

From the first equation,

2x+ 3*10 =40

2x = 40-30

= 10

x = 5

Answer

:

Slide11

Now consider another problem,

“ The price of 3 pencils and 4 pens is 26 rupees; and for 6 pencils and 3 pens it is 27 rupees. What is the price of each?”

What

is the difference of this problem with

the previous

ones?

Slide12

Let’s write the prices like this:

The number of pencils in two rows are different. How can we make them same? Think.

Answer

:

PENCIL

PEN

PRICE

3

4

26

6

3

27

Slide13

We can redraw the table as follows by multiplying the first row by 2 to make the number of pencils in both cases are same.

Now in the case of algebra this table becomes,

Let the price of a pencil = x & that of a pen = y. then we have,

3x+4y = 26

6x+3y = 27

PENCIL

PEN

PRICE

3

4

26

6

3

27

6

8

52

Slide14

Multiplying the first equation by 2 we get,

2(3x+4y=26) ⇨ 2(3x+4y) =52

⇨ 6x+8y = 52

Then by solving the above equation and the second equation above obtained we get

(6x+8y)-(6x+3y) = 52-27

8y-3y = 25

5y = 25

Slide15

y = 25/5

=5

Now substitute the value of y in first equation, we get,

3x+(4*5) = 26

3x+20 = 26

3x = 6

x = 2

Slide16

Consider another problem,

“Five small buckets and two small buckets of water make 20 litres; two small buckets and 3 large buckets make only 19 litres. How much water can each bucket hold?”

Slide17

Answer:

Let the capacity of small bucket = x litres

& the capacity of large bucket = y litres

We can write the given facts as below,

5x+2y = 20 →(1)

2x+3y = 19 →(2)

We can change the equations as,

* 2

:

10x+4y = 40

* 5

: 10x+15y = 95

Slide18

Now subtracting the above two equations we get,

11y = 55

y

= 5

By putting the value of y in (1) we get,

5x+10 = 20

5x = 10

x =

2

Thus the small bucket can hold 2 litres and the large bucket 5 litres.

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SOME OTHER EQUATIONS

Consider the problem,

“ Of two squares, sides of the larger are 5 cm longer than the smaller and the area of the larger is 55cm more. What is the length of the sides of each?”

Slide20

Answer:

Take the length of a side of large square as x centimetres and that of smaller as y centimetres, we can put the facts as two equations:

x-y = 5

x

2

–

y

2

= 55

Now what can we do?

Slide21

Here we use the identity,

x

2

–

y

2

= (x+y)(x-y)

So in this problem the equations become;

x-y =5

(x+y)(x-y) = 55

On dividing the two equations we get,

x+y = 11

Now consider the two equations,

Slide22

x+y = 11

x-y = 5

By solving this we get,

2x

= 11+5

=16

x= 8

2y

=

11-5

= 6

y = 3

Thus the length of sides of squares are 8cm and 3cm.

Slide23

Consider the problem,

“ Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be doubled. What are their ages now?”

Slide24

Answer:

Let present age of Rahim = x

& present age of Ramu =y

Before 4 years,

x

-4 = 3(y-4)

x-4 = 3y-12

x-3y = -8 → (1)

After 2 years,

x+2 = 2(y+2)

x+2 = 2y+4

x-2y = 2 →(2)

Slide25

By subtracting (2) from (1)

we get,

-y = -10

y = 10

By putting the value of y in (1) we get,

x-(3*10) = -8

x = -8+30

= 22.

So present age of Rahim = 22 & Ramu = 10

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THANKYOU

PREPARED BY,

TELMA JOBY

NO: 24