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Question
Solve by the method of elimination
3(2x + y) = 7xy, 3(x + 3y) = 11xy
Solution
3(2x + y) = 7xy
6x + 3y = 7xy
Divided by xy
`(6x)/(xy) + (3y)/(xy) = (7xy)/(xy)`
`6/y + 3/x` = 7
Let `1/x` = a, `1/y` = b
3a + 6b = 7 → (1)
3(x + 3y) = 11xy
3x + 9y = 11xy
Divided by xy
`(3x)/(xy) + (9y)/(xy) = (11xy)/(xy)`
`3/y + 9/x` = 11
Let `1/x` = a, `1/y` = b
9a + 3b = 11 → (2)
(1) × 3 ⇒ 9a + 18b = 21 → (3)
(2) × 1 ⇒ 9a + 3b = −11 → (2)
(3) – (2) ⇒ 15b = 10
b = `10/15 = 2/3`
Substitute the value of b = `2/3` in (1)
`3"a" + 6 xx 2/3` = 7
3a + 4 = 7
3a = 7 – 4
3a = 3
a = `3/3`
= 1
But `1/x` = a
`1/x` = 1
x = 1
But `1/y` = b
`1/y = 2/3`
2y = 3
y = `3/2`
∴ The value of x = 1 and y = `3/2`
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