SOLVING SYSTEMS OF EQUATIONS BY ELIMINATION AND SUBSTITUTION EXAMPLES

Solved Examples

Example 1 :

Solve the following pairs of linear equations by the elimination method and the substitution method

(i) x + y = 5 and 2 x – 3 y = 4

Solution :

x + y = 5 --------(1)

2 x – 3 y = 4 --------(2)

Elimination method

(1) ⋅ 3 ==> 3x + 3y  =  15

(1) + (2)  2x - 3y  =  4

            -------------------

        5x  =  19 

   x  =  19/5

By applying x  =  19/5 in (1) equation, we get

(19/5) + y  =  5

y  =  5 - (19/5)

y  =  (25 – 19)/5

y  =  6/5

Therefore solution is x  =  19/5 and y  =  6/5

Now, let us do the same problem in substitution method :

Substitution method

x + y = 5 --------(1)

2 x – 3 y = 4 --------(2)

Step 1 :

Find the value of one variable in terms of another variable

y  =  5 – x

Step 2 :

Substitute this value of y in the other equation, and reduce it to an equation in one variable.

2x – 3(5 – x)  =  4

2x – 15 + 3 x  =  4

5x  =  4 – 15

5x  =  19

x  =  19/5

Step 3 :

Apply x  =  19/5 in the equation y = 5 – x

y  =  5 – (19/5)

y  =  (25 – 19)/5

 y  =  6/5

So, the solution is x  =  19/5 and y  =  6/5

Example 2 :

Solve the following pairs of linear equations by the elimination method and the substitution method

3x + 4y  =  10 and 2x – 2y  =  2

Solution :

3x + 4y  =  10 ---------(1)

 2x – 2y  =  2 ---------(2)

Elimination method

(1) + (2)    3x + 4y  =  10

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

               ----------------

                  7x  =  14

                  x  =  14/7  =  2

By applying the value of x in (2), we get

2(2) – 2y  =  2

4 – 2y  =  2

-2y  =  2 – 4

 -2y  =  -2

y  =  1

Therefore solution is x  =  2 and y  =  1

Substitution method

3 x + 4 y = 10 ---------(1)

 2 x – 2 y = 2 ---------(2)

Step 1 :

Find the value of one variable in terms of another variable

-2y  =  2 – 2 x

y  =  (2 x – 2)/2

y  =  x - 1

Step 2 :

Substitute this value of y in the other equation and reduce it to an equation in one variable.

3x + 4(x - 1)  =  10

3x + 4x – 4  =  10

7x – 4  =  10

7x  =  10 + 4

7x  =  14

x  =  2

Step 3 :

Apply x = 2 in the equation y = x - 1

y = 1

So, the solution is x = 2 and y = 1.

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