# 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

#### 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?

#### 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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