Question

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Answer 1

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Answer 2

Question:

Solve the given system of equations:

21/x + 47/y = 110

47/x + 21/y = 162 , x,y ≠ 0

Answer:

x = 1/3 , y = 1

Solution:

The given system of equation are :

21/x + 47/y = 110 -------(1)

47/x + 21/y = 162 -------(2)

Now,

Let 1/x = a and 1/y = b

Also,

Putting 1/x = a and 1/y = b in eq-(1) and eq-(2) ,

We have ;

21a + 47b = 110 ------(3)

47a + 21b = 162 ------(4)

Now,

Adding eq-(1) and eq-(2) , we get ;

=> 21a + 47b + 47a + 21b = 110 + 162

=> 68a + 68b = 272

=> 68(a + b) = 272

=> a + b = 272/68

=> a + b = 4 --------(5)

Now,

Subtracting eq-(1) from eq-(2) , we get ;

=> (47a + 21b) - (21a + 47b) = 162 - 110

=> 47a + 21b - 21a - 47b = 162 - 110

=> 26a - 26b = 52

=> 26(a-b) = 52

=> a - b = 52/26

=> a - b = 2 --------(6)

Now,

Adding eq-(5) and eq-(6) , we get ;

=> a + b + a - b = 4 + 2

=> 2a = 6

=> a = 6/2

=> a = 3

=> 1/x = 3

=> x = 1/3

Now,

Putting a = 3 , in eq-(5) , we get ;

=> a + b = 4

=> 3 + b = 4

=> b = 4 - 3

=> b = 1

=> 1/y = 1

=> y = 1

Hence,

The solution of the given system of equations is;

x = 1/3 , y = 1