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Solve the Following Systems of Equations: `2/X + 5/Y = 1` `60/X + 40/Y = 19, X = ! 0, Y != 0` - Mathematics

Solve the following systems of equations:
`2/x + 5/y = 1`

`60/x + 40/y = 19, x = ! 0, y != 0`

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Solution

Taking 1/x = u and 1/y = v the given become

2u + 5v = 1  ......(1)

60u + 40u = 19 .....(ii)

Let us eliminate ‘u’ from equation (i) and (ii), multiplying equation (i) by 60 and equation (ii) by 2, we get

120u + 300v = 60 ....(iii)

120u + 80v = 38 ....(iv)

Subtracting (iv) from (iii), we get

300v - 80v = 60 - 38

=> 220v = 22

=> `v = 22/220 = 1/10`

Putting `v = 1/10` in equation (i) we get

`2u + 5 xx 1/10 = 1`

`=> 2u + 1/2 = 1`

`=> 2u = 1 - 1/2`

`=> 2u = (2-1)/2 = 1/2`

`=> 2u = 1/2`

`=> u = 1/4`

hence `x = 1/u  = 4` and `y = 1/v = 10`

So, the solution of the given system of equation is x = 4, y = 10

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.3 | Q 20 | Page 45
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