# Solve the Following Systems of Equations: 2/X + 3/Y = 9/(Xy) 4/X + 9/Y = 21/(Xy), X != 0, Y != 0 - Mathematics

Solve the following systems of equations:

2/x + 3/y = 9/(xy)

4/x + 9/y = 21/(xy), where x != 0, y != 0

#### 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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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Q 22