## ELIMINATION METHOD WORKSHEET

Questions 1-8 : Solve each system by elimination.

Question 1 :

x + 2y = 7

x – 2y = 1

Question 2 :

3x + y = 8

5x + y = 10

Question 3 :

x + ʸ⁄ = 4

ˣ⁄₃ + 2y = 5

Question 4 :

11x - 7y = xy

9x - 4y = 6xy

Question 5 :

³⁄y + ⁵⁄ₓ = ²⁰⁄ₓy

2⁄ₓ + ⁵⁄y = 15⁄ₓy

Question 6 :

8x – 3y = 5xy

6x – 5y = -2xy

Question 7 :

13x + 11y = 70

11x + 13y = 74

Question 8 :

65x – 33y = 97

33x – 65y = 1

x + 2y = 7 ----(1)

x – 2y = 1 ----(2)

(1) + (2) :

Divide both sides by 2.

x = 4

Substitute x = 8 into (1).

4 + 2y = 7

Subtract 4 from both sides.

2y = 3

Divide both sides by 2.

y = ³⁄₂

Thereforem the solution is

(x, y) = (4, ³⁄₂)

3x + y = 8 ----(1)

5x + y = 10 ----(2)

(2) - (1) :

Divide both sides by 2.

x = 4

Substitute x = 8 into (1).

4 + 2y = 7

Subtract 4 from both sides.

2y = 3

Divide both sides by 2.

y = ³⁄₂

Thereforem the solution is

(x, y) = (4, ³⁄₂)

x + ʸ⁄ = 4 ----(1)

ˣ⁄₃ + 2y = 5 ----(2)

Multiply (1) by 2.

2x + y = 8 ----(3)

Multiply (2) by 3.

x + 6y = 15 ----(4)

2(4) - (3) :

Divide both sides by 11.

y = 2

Substitute y = 2 into (4).

x + 6(2) = 15

x + 12 = 15

Subtract 12 from both sides.

x = 3

Thereforem the solution is

(x, y) = (3, 2)

11x - 7y = xy ----(1)

9x - 4y = 6xy ----(2)

Divide both sides of (1) by xy.

¹¹⁄y - ⁷⁄ₓ = 1

9⁄y - 4⁄ₓ = 6

Let a = 1⁄ₓ and b = 1⁄y.

Then, we have

11b + 7a = 1 ----(3)

9b - 4a = 6 ----(4)

9(3) - 11(4) :

Divide both sides by -19.

a = 3

Substitute a = 3 into (4).

9b - 4(3) = 6

9b - 12 = 6

9b = 18

Divide both sides by 2.

b = 2

 a = 31⁄ₓ = 3x = ⅓ b = 21⁄y = 2y = ½

Therefore, the solution is

(x, y) = (, ½)

³⁄y + ⁵⁄ₓ = ²⁰⁄ₓy ----(1)

2⁄ₓ + ⁵⁄y = 15⁄ₓy ----(2)

Multiply both sides of (1) by xy.

3x + 5y = 20 ----(3)

Multiply both sides of (2) by xy.

2x + 5y = 15 ----(4)

(3) - (4) :

(3x + 5y) - (2x + 5y) = 20 - 15

3x + 5y - 2x - 5y = 5

x = 5

Substitute x = 5 into (4).

2(5) + 5y = 15

10 + 5y = 15

Subtract 10 from both sides.

5y = 5

Divide both sides by 5.

y = 1

Therefore, the solution is

(x, y) = (5, 1)

8x – 3y = 5xy ----(1)

6x – 5y = -2xy ----(2)

Divide both sides of (1) by xy.

8⁄y - 3⁄ₓ = 5

6⁄y - 5⁄ₓ = -2

Let a = 1⁄ₓ and b = 1⁄y.

Then, we have

8b - 3a = 5 ----(3)

6b - 5a = -2 ----(4)

5(3) - 3(4) :

5(8b - 3a) - 3(6b - 5a) = 5(5) - 3(-2)

40b - 15a - 18b + 15a = 25 + 6

22b = 31

Divide botyh sides by 22.

b = ³¹⁄₂₂

Substitute b³¹⁄₂₂ into (4).

8(³¹⁄₂₂) - 3a = 5

¹²⁴⁄₁₁ - 3a = 5

Multiply both sides by 11.

124 - 33a = 55

Subtract 124 from both sides.

-33a = -69

Divide both sides by -33.

a = ²³⁄₁₁

 a = ²³⁄₁₁1⁄ₓ = ²³⁄₁₁x = ¹¹⁄₂₃ b = ³¹⁄₂₂1⁄y = ³¹⁄₂₂y = ²²⁄₃₁

Therefore, the solution is

(x, y) = (¹¹⁄₂₃, ²²⁄₃₁)

13x + 11y = 70 ----(1)

11x + 13y = 74 ----(2)

coefficient of x in (1) = coefficient of y in (2)

coefficient of y in (1) = coefficient of x in (2)

(1) + (2) :

24x + 24y = 144

Divide both sides by 24.

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

(1) - (2) :

2x – 2y = -4

Divide both sides by 2.

x – y = -2 ----(4)

(3) + (4) :

2x = 4

x = 2

Substitute x = 2 into (3).

2 + y = 6

y = 4

Therefore, the solution is

(x, y) = (2, 4)

65x – 33y = 97 ----(1)

33x – 65y = 1 ----(2)

coefficient of x in (1) = coefficient of y in (2)

coefficient of y in (1) = coefficient of x in (2)

(1) + (2) :

98x - 98y = 98

Divide both sides by 98.

x - y = 1 ----(3)

(1) - (2) :

32x + 32y = 96

Divide both sides by 32.

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

(3) + (4) :

2x = 4

x = 2

Substitute x = 2 into (4).

2 + y = 3

y = 1

Therefore, the solution is

(x, y) = (2, 1)

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