Innovative Lesson

Mathematics Class  9(part- I) 04-Jun-16 Minu Michael

unit -3

Pair of equations

Mental math and algebra

Let’s start with a problem;

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

  We can do this in several ways. Taking out 10 extra black balls for the time being, we have 90 left in the box, black and white are equal. So 45 of each. Adding the 10 black beads kept apart, black becomes 55 and white remains 45.

  We can do it with algebra also.(Recall the chapter Equations in class 8).

Taking the number of black balls as ‘ x ’,the number of white balls is ‘ x-10 ‘ and since there are 100 in all,

                                                  x + (x-10) = 100

We can find out x from this;

                                                   2x – 10 = 100

                                                   2x = 100 +10

                                                  2x = 110

                                                     x = 55.

Thus we find the number of black balls as 55; subtracting 10, we get the number of white balls as 45.

There is another way of using algebra again. Taking number of black balls as ‘x’ and the number of white balls as ‘y’, we can write what we are told as two equations.

                                 x + y = 100

                                 x – y = 10

How do we separate x and y from these?

Recall what we learnt about the sum and difference of two numbers, in class 7; adding the sum and difference of two numbers gives twice the larger number.

(The section, sum and difference in the lesson, Unchanging Relation) .

And we also saw that the difference subtracted from the sum gives twice the smaller number.

So in our problem, 

                                   2x =  110

                            Þ    x = 55

So the value of y can find out by putting the value of x in any equation.

ie,

                 x + y = 100

        Þ  55 + y =100

       Þ   y = 100 – 55

                 =45.

There is another problem;

                                            The price of a table and chair and table together is 5000rupees. The price of a table and four chairs is 8000 rupees. What is the price of each?

Now we can solve it by using algebra as follows:

Firstly taking the price of a table as x and so the price of chair is 5000-x( sine price of both is 5000).Now the price of a table and 4 chairs is 8000. So in equation form

                                x + 4 (5000-x) = 8000.

Now to solve to find out the value of x .

 ie,                        x + (4*5000) – (4*x) =8000

                             x + 20000 – 4x = 8000

                                 -3x = 8000 - 20000

                                 -3x = -12000

                                   x = -12000/-3

                                     =4000.

So the value of one table is 4000. So the value of one chair is 1000 (Since 5000 in both).

Solve it taking price of chair as x and price of table as y.

                              Then, x + y = 5000

                                         x + 4y = 8000

By solving these two equations we get,

                                                    3y = 3000

                                                      y=1000.

That is the price of each chair is 1000 and price of one table is 4000 ( since the sum of both is 5000).

One more interesting problem,

                                                 When we add one to the numerator of a fraction and simplify it, we get ½. When we add one to the denominator instead and simplify it, we get 1/3. What is this fraction?

Let’s the fraction be x/y  .

                            So, ( 1+x) / y = ½

                                    x/ (1+y) = 1/3

We can solve it by cross multiplying both eqations ( the lesson, Fractions)      

 So,                                 2(1+x) = y

1

                                       2 +2x = y

                                ie, 2x – y = -2

2

The second fraction on cross multiplying becomes,

                                       3x = y+1

On solving 1 and 2 we get,

                                x = 3, and y = 8.

So the fraction is 3/8.

Now to solve some interesting problems like this:

1). In a rectangle of perimeter one meter, one side is five centimeters longest than the other. What are the lengths of the sides?

                                  

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

3). A man invested 10000rupees, split into two schemes, at annual rates of interests 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invested each?

4). A three and half metres long rod is to be cut into two pieces, one piece  is to be bent into a square and the other into an equilateral triangle. The length of a side of both must be same. How should it be cut?

Two Equatiion

 See this problem:-

The price of three pen and four note books is 40 rupees; and the price of two pens and five notebooks is 60 rupees. What are the prices of a pen and notebook?

        We can solve the problem by using sufficient variables. now taking the price of a pen be x and price of a notebook be y, then the corresponding equation be

2x+3y=40

2x+5y=60

On solving these two equations we get,

2y=20

ie, y=10

Now substitute the value of y in any of these equations gives the value of x.

2x+3y=40                                   2x + 10= 40

                                                        2x = 30

                                                         x = 15

Now let’s look on another problem:-

The price of 3 apples and 4 oranges is 26 rupees; and for 6 apples and 3 oranges, it is 27 rupees. What are the prices of each?

1

Let the price of an apple be x and price of an orange be y, then the equations are

3x + 4y=26

2

6x + 3y=27

3

1

Here we can’t solve these equations directly, so in order to cancel the variable multiply           with 2, we get

3

2

6x + 8y = 52

Now                  and              together gives the solution,

ie , 5y = 25

y = 5

Now 3x +4y =26 becomes, 3x + 20 = 26’

ie, 3x = 6

x = 2

Illustrated in a simple way,

Coefficient of x	Coefficient of y	Real number
3	4	26
6	3	27

After making coefficients of one variable alike it becomes:

Coefficient of x	Coefficient of y	Real number
6	8	52
6	3	27

So by direct subtraction it becomes,

5y = 25                         y = 5  and  x =  2.

Help Ammu…..

One day Ammu asked to her mother that how old she was? Mother replied that 4 yours ago my age is thrice your age and 2 years later my age become double of you, after few minutes Ammu correctly tells her mother’s age. Then what is the answer?

Is men/women are best?

        Only 4 men and 3 women can complete a work in 13 days. The same work can be completed by 8 men and 2 women in 10 days. Then how many days are needed for men and women only?