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.
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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.
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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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