Solve the following systems of equations:
`2/x + 3/y = 9/(xy)`
`4/x + 9/y = 21/(xy), where x != 0, y != 0`
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Solution
The system of given equation is
`2/x + 3/y = 9/(xy)` ....(i)
`4/x + 9/y = 21/(xy), x != 0, y != 0` ......(ii)
Multiplying equation (i) adding equation (ii) by ,xy we get
2y + 3x = 9 ....(iii)
4y + 9x = 21 .....(iv)
From (iii), we get
3x = 9 - 2y
`=> x = (9 - 2y)/3`
Substituting x `= (9 - 2y)/3` in equation (iv) weget
`4x + 9((9 - 2y)/3) = 21`
=> 4y + 3(9 - 2y) = 21
=> 4y + 27 - 6y = 21
`=> -2y = 21 - 27`
=> -2y = -6
`=> y = (-6)/(-2) = 3`
Putting y = 3 in x = (9 - 2y)/3 we get
`x = (9 - 2xx3)/3`
`= (9-6)/3`
= 3/3
= 1
Hence, solution of the system of equation is x = 1, y = 3
Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method
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