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')