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 = ⅓
Substitute a = ⅓ into (2).
6(⅓)  3b = 1
2  3b = 1
Subtract 2 from both sides.
3b = 1
Divide both sides by 3.
b = ⅓
a = ⅓ x  1 = 3 x = 4 
b = ⅓ y  2 = 3 y = 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 = 1 ¹⁄ₓ = 1 x = 1 
b = 1 ¹⁄y = 1 y = 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 = 1 ¹⁄ₓ = 1 x = 1 
b = ½ ¹⁄y = ½ y = 2 
Therefore, the solution is
(x, y) = (1, 2)
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