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

Is there an error in this question or solution?

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