Class 10 : Exercise 3.3

1 . Solve the following pair of linear equations by the substitution method

i) x + y = 14
   x - y = 4

Solution

Lets name the given equations are

   x + y = 14    -----------------> (1)

   x - y = 4      ------------------> (2)

we have to solve the given equations using substitution method

Step 1:

Consider any one of the equation and isolate any one of the variable

here, lets take the first equation and isolate the variable 'x'

x + y = 14    -----------------> (1)

to move y variable to another side we have to  subtract y on both sides

 x + y = 14

     - y    -y    

 x       = 14-y 

now lets name the above equation as (3)

so, x = 14 - y   -------------> (3)

Step 2:

Now substitute equation (3) in another given equation

here, lets substitute equation 3 in 2

that is, 

         x - y = 4      ------------------> (2)

(14 -y ) - y = 4

14 -y -y = 4

14 - 2y = 4 

-14       = -14    ( subtract 14  on both sides )

 0  - 2y  = -10

      

-2y = -10

divide both sides by -2

 -2y = -10

 -2       -2

y = 5

Step 3:

Substitute y variable in any one of the given equations

lets take equation (1) and substitute y = 5

x + y = 14    -----------------> (1)

x + 5 =14   ( to move the constant 5 to another side we have to subtract 5 on both sides)

   - 5 = -5  

x + 0 =  9 

hence

 x = 9 and y = 5

ii) s - t = 3

   (s/3) + (t/2) = 6

Solution 

Lets name the given equations are

               s - t = 3 -----------> (1)

   (s/3) + (t/2) = 6 -----------> (2)

Step 1

Consider any one of the equation and isolate any one of the variable

here, lets take the first equation and isolate the variable 's'

        s - t = 3  ------------> (1)

to move t variable to another side we have to add t on both sides

  s - t = 3

    +t   +t    

 s       = 3+t 

now lets name the above equation as (3)

s = 3 + t ----------> (3)

Step 2:

Now substitute equation (3) in another given equation

here, lets substitute equation 3 in 2

that is, 

(s/3) + (t/2) = 6 -----------> (2)

(3+t)/3 + (t/2) =6

3/3 + t/3 + t/2 = 6

1 + t/3 + t/2 = 6  (to move 1 to right side,we have to subtract 1 on both sides) 

so, t/3 + t/2 = 6-1

     t/3 + t/2 = 5   ( Lcm of 3 and 2 = 6)

that is, 

     ( 2t + 3t ) / 6 = 5 

        5t / 6 = 5    ( multiply by 6 on both sides )

         5t = 5*6

         5t = 30  ( to find t we have to divide both sides by 5 )

          5      5

          t = 6

Step 3:

Substitute t variable in any one of the given equations

lets take equation (1) and substitute t = 6

  s - t = 3 -----------> (1)

  s - 6 = 3

    + 6   +6

  s      = 9  

Hence

 s = 9 and t =6


iii)  3x - y = 3
      9x - 3y = 9 

Solution

Lets name the given equations as

      3x - y = 3    ------------> (1)
      9x - 3y = 9  ------------> (2)

Step 1

Consider any one of the equation and isolate any one of the variable

here, lets take the first equation and isolate the variable 'x'

 3x - y = 3    ------------> (1)
    + y = +y   (to move y to right side we just added y on both sides)

 3x    = 3 + y
 to make 3x as 3 we have to divide both sides by 3
that is,
 3x  = (3 + y) 
  3          3

x = (3+y) / 3 -----------> (3)

Step 2:

Now substitute equation (3) in another given equation

here, lets substitute equation 3 in 2

that is, 
     9x - 3y = 9  ------------> (2)

    3 9(3 + y)  - 3y = 9

          3
        3(3+y) -3y = 9
          9 + 3y -3y = 9
          9  + 0 = 9
              9  =  9 ( which is always true )

Hence, the given pair of equations has infinite possible number of solutions.

Note :

When we try to solve both equations if we get any number = same number
then that equation has infinite number of solutions
 
When we try to solve both equations if we get any number != same number
then that equation has NO solutions 
(!= meaning 'not equal')
 

Class 10 : Exercise 3.1

1. Aftab tells his daughter , "seven years ago, i was seven times as old as you were. Also three years from , I shall be three times as old as you will be ." Represent this situation algebraically and graphically . 

Solution

Let us assume, the current age of Aftab = x
                         the current age of his daughter = y

 Seven years ago ,
                      the age of Aftab = x-7
                      the age of his daughter = y-7
Given,
   seven years ago, i was seven times as old as you were
that is,
       x-7 = 7(y-7)
       x - 7 = 7y -49
 now lets keep x and y on one side
that is, lets move y to the left side
to move 7y to the  left side, we have to subtract 7y on both sides
so the above equation becomes
       x-7y - 7 = -49
now lets move the constant to the right side
so, to move -7 to right side, we have to add 7 on both sides
which becomes,
       x - 7y = -49 + 7
       x - 7y = -42 
lets name the above equation as (1)

       x - 7y = -42 -----------------> ( 1 )

After 3 years ,  the age of Aftab = x + 3
                         the age of his daughter = y + 3
Also given , three years from , I shall be three times as old as you will be
that is ,    x + 3 = 3 ( y + 3 )
                x + 3 = 3y + 9
lets keep variables on left side and constant on right side
so to move 3y to left side, we have to subtract 3y on both sides 
                x - 3y + 3 = 9
                x - 3y = 9 - 3
                x - 3y = 6 
lets name the second equation as ( 2 )

                x - 3y = 6 ----------------> ( 2 )

Therefore the algebraic equations are

            x - 7y = -42 -----------------> ( 1 )
            x - 3y = 6    ---------------->   ( 2 )

Now , lets represent the above equations graphically

Lets take the first equation
        x - 7y = -42
lets do the table
lets assume x values as -7 , 0 , 7 and find its corresponding y values by substituting x values in the above equation

      x    -7     0     7 

      y      5     6     7        

Lets take the second equation
          x - 3y = 6
lets do the table
lets assume x values as -3 , 0 , 3 and find its corresponding y values by substituting x values in the above equation

     x    -3     0     3 

     y    -3     -2     -1       






2 . The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later she buys another bat and 3 more balls of the same kind for Rs 1300 . Represent this situation algebraically and graphically . 

Solution

let us assume,
       Cost of a bat = x
       Cost of a ball = y
Given,
       cricket team buys 3 bats and 6 balls for Rs 3900
which means
       3x + 6y = 3900   -------------->  (1)
also given,
       she buys another bat and 3 more balls of the same kind for Rs 1300
that is
       x + 3y = 1300  -----------------> (2)

Therefore the algebraic equations are 

            3x + 6y = 3900   -------------->  (1)
            x + 3y = 1300  -----------------> (2)

Now lets represent the above equations graphically

Lets take our first equation

           3x + 6y = 3900   -------------->  (1)


lets do the table
lets assume x values as 100 , 0 , -100 and find its corresponding y values by substituting x values in the above equation

x     100     0     -100

y     600    650   700

Lets take the second equation

           x + 3y = 1300  -----------------> (2)

lets do the table
lets assume x values as 400 , 100 , -200 and find its corresponding y values by substituting x values in the above equation

x     400    100    -200

y     300    400     500 

Lets do the graph
 

3 . The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160 . After a month , the cost of 4 kg of apples and 2 kg of grapes are Rs 300 . Represent this situation algebraically and graphically.  

Solution

lets say,
   Cost of one kg of apples = x
   Cost of one kg of grapes =  y

Given,
The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160 
that can be written as 

2x + 1y = 160  -------------------> (1)

also given,
the cost of 4 kg of apples and 2 kg of grapes are Rs 300
which is

4x + 2y = 300 -------------------> (2)

Therefore the algebraic equations are 

 
2x + 1y = 160  -------------------> (1)
4x + 2y = 300 -------------------> (2)

Now lets represent the above equations graphically

Lets take our first equation

         2x + 1y = 160 -------------->  (1)

lets do the table
lets assume x values as 50 , 60 , 70 and find its corresponding y values by substituting x values in the above equation

x     50     60    70

y     60     40    20 

Lets take the second equation

          4x + 2y = 300  -----------------> (2)

lets do the table
lets assume x values as 400 , 100 , -200 and find its corresponding y values by substituting x values in the above equation

x    30    40    50

y    90    70    50

Lets do the graph