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Question
Solve by the method of elimination
`4/x + 5y` = 7, `3/x + 4y` = 5
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Solution
Let `1/x` = a
4a + 5y = 7 → (1)
3a + 4y = 5 → (2)
(1) × 4 ⇒ 16a + 20y = 28 →(3)
(2) × 5 ⇒ 15a + 20y = 25 → (4)
(3) – (4) ⇒ a + 0 = 3
a = 3
Substitute the value of a = 3 in (1)
4(3) + 5y = 7
5y = 7 – 12
5y = −5
5y = `(-5)/5` = −1
But `1/x` = a
`1/x` = 3
3x = 1 ⇒ x = `1/3`
The value of x = `1/3` and y = −1
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