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Solve the Following Systems of Equations: `2x - 3/Y = 9` `3x + 7/Y = 2, Y != 0` - Mathematics

Solve the following systems of equations:

`2x - 3/y = 9`

`3x + 7/y = 2,  y != 0`

 

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Solution

The given systems of equation is

`2x - 3/y = 9`  .....(i)

`3x + 7/y = 2, y != 0 `   ...(ii)

Taking `1/y = u` the given equations becomes

2x - 3u = 9 ....(iii)

3x + 7u = 2 ....(iv)

From (iii), we get

2x = 9 + 3u

`=> x = (9 + 3u)/2`

Substituting `x = (9 + 3u)/2` in iv we get

`3((9 + 3u)/2) + 7u = 2`

`=> (27 + 9u + 14u)/2 = 2`

`=> 27 + 23u =  2xx2`

`=> 23u = 4 - 27`

`=> u = (-23)/23 = -1`

hence `y = 1/u = 1/(-1) = -1`

Putting u = -1 in `x = (9 + 3u)/2`  we get

`x = (9 + 3(-1))/2 = (9-3)/2 = 6/2 = 3`

=> x = 3

Hence, solution of the given system of equation is x = 3, y = -1

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.3 | Q 13 | Page 44
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