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प्रश्न
Solve by the method of elimination
`4/x + 5y` = 7, `3/x + 4y` = 5
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उत्तर
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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* Question modified
Samidha and Shreya have pocket money Rs.x and Rs.y respectively at the beginning of a week. They both spend money throughout the week. At the end of the week, Samidha spends Rs.500 and is left with as much money as Shreya had in the beginning of the week. Shreya spends Rs.500 and is left with `(3)/(5)` of what Samidha had in the beginning of the week. Find their pocket money.
Solve by the method of elimination
3(2x + y) = 7xy, 3(x + 3y) = 11xy
