SOLVING EQUATIONS USING ELIMINATION AND SUBSTITUTION

Example 1 :

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

3x – 5y – 4 = 0 and 9x = 2y + 7

Solution :

3x – 5y = 4 ----(1)

 9x - 2y = 7----(2)

Elimination Method :

Multiply (1) by -3 and it to (2) to eliminate x.

(1) ⋅ -3 + (2) :

13y = -5

y = -5/13

Substitute y = -5/13 in (1).

3x – 5(-5/13) = 4

3x + (25/13) = 4

(39x + 25)/13 = 4

39x + 25 = 4(13)

39x + 25 = 52

39x = 52 – 25

39x = 27

x = 27/39

x = 9/13

So, the solution is (x, y) = (9/13, -5/13).

Substitution Method :

3x – 5y = 4 ----(1)

 9x - 2y = 7 ----(2)

Step 1 :

Find the value of one variable in terms of another variable.

In (2), solve for y in terms of x.

-2y = 7 – 9x

y = (9x – 7)/2

Step 2 :

Substitute y = (9x - 7)/2 in (1).

3x – 5(9x – 7)/2 = 4

(6x - 45x + 35)/2 = 4

-39x + 35 = 4(2)

-39x = 8 - 35

-39x = -27

x = -27/(-39)

x = 9/13

Step 3 :

Substitute x = 9/13 in y = (9 x – 7)/2.

y = [9(9/13) – 7]/2

y = [(81/13) – 7]/2

y = [(81 –91)/26]

 y = -10/26

y = -5/13

So, the solution is (x, y) = (9/13, -5/13).

Example 2 :

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

x/2 + 2y/3 = -1 and x – y/3 = 3

Solution :

(3x + 4y)/6 = -1 ----> 3x + 4y = -6 ----(1)

x – y/3 = 3 ----> 3x – y = 9 ----(2)

Elimination Method :

Subtract (2) from (1) to eliminate x.

5y = -15

y = -3

Substitute y = -3 in (2).

3x - (-3) = 9

3x + 3 = 9

3x = 6

x = 2

So, the solution is (x, y) = (2, -3).

Substitution Method :

3x + 4y = -6 ----(1)

3x – y = 9 ----(2)

Step 1 :

Find the value of one variable in terms of another variable

Solve for y in terms of x in (2).

-y = 9 – 3x

 y = 3x - 9

Step 2 :

Substitute y = 3x - 9 in (1).

3x + 4(3x - 9) = -6

3x + 12x – 36 = -6

15x – 36 = -6

15x = -6 + 36

15x = 30

x = 2

Step 3 :

Substitute x = 2 in y = 3x - 9.

y = 3(2) – 9

y = 6 - 9

y = -3

So, the solution is (x, y) = (2, -3).

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