# SOLVING DIFFICULT PROBLEMS USING ELIMINATION METHOD

Solve the each of the following pairs of equations using elimination method.

Example 1 :

Solution :.

Then, we have

5a + b = 2 ----(1)

6a – 3b = 1 ----(2)

3(1) + (2) :

3(5a + b) + (6a - 3b) = 3(2) + 1

15a + 3b + 6a - 3b = 6 + 1

21a = 21

Divide each side by 21.

a =

Substitute a =  into (2).

6() - 3b = 1

2 - 3b = 1

Subtract 2 from both sides.

-3b = -1

Divide both sides by -3.

b =

 a = ⅓ x - 1 = 3x = 4 b = ⅓ y - 2 = 3y = 5

Therefore, the solution is

(x, y) = (4, 5)

Example 2 :

7x - 2y = 5xy

8x + 7y = 15xy

Solution :

Divide both sides of the given equations by xy.

Let a = ¹⁄ₓ and b = ¹⁄y.

Then, we have

7b - 2a = 5 ----(1)

8b + 7a = 15 ----(2)

7(1) + 2(2) :

7(7b - 2a) + 2(8b + 7a) = 7(5) + 2(15)

49b - 14a + 16b + 14a = 35 + 30

65b = 65

Divide both sides by 65.

b = 1

Substitute b = 1 into (1).

7(1) - 2a = 5

7 - 2a = 5

Subtract 7 from both sides.

-2a = -2

Divide both sides by -2.

a = 1

 a = 1¹⁄ₓ = 1x  =  1 b =  1¹⁄y = 1y = 1

Therefore, the solution is

(x, y) = (1, 1)

Example 3 :

6x + 3y = 6xy

2x + 4y = 5xy

Solution :

Divide both sides of the first equation by 3xy and the second equation by xy.

Let a = ¹⁄ₓ and b = ¹⁄y.

Then, we have

2b + a = 2 ----(1)

2b + 4a = 5 ----(2)

(2) - (1) :

(2b + 4a) - (2b + a) = 5 - 2

2b + 4a - 2b - a = 3

3a = 3

Divide both sides by 3.

a = 1

Substitute a = 1 into (1).

2b + 1 = 2

Subtract 1 from both sides.

2b = 1

Divide both sides by 2.

b = ½

 a = 1¹⁄ₓ = 1x  =  1 b = ½¹⁄y = ½y = 2

Therefore, the solution is

(x, y) = (1, 2)

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