Question

solve the equation 2/x-3/y+3/z=10 , 1/x+1/y+1/z=10, 3/x-1/y+2/z=13​

Answer 1

Answer :

x = 1/2 , y = 1 , z = 1/7

Solution :

Here ,

The given equations are ;

2/x - 3/y + 3/z = 10

1/x + 1/y + 1/z = 10

3/x - 1/y + 2/z = 13

Putting 1/x = X , 1/y = Y , 1/z = Z , the given equations will become ;

2X - 3Y + 3Z = 10 ---------(1)

X + Y + Z = 10 --------(2)

3X - Y + 2Z = 13 ----------(3)

Now ,

Adding eq-(1) and 3×(2) , we get ;

=> 2X - 3Y + 3Z + 3×(X + Y + Z) = 10 + 3×10

=> 2X - 3Y + 3Z + 3X + 3Y + 3Z = 10 + 30

=> 5X + 6Z = 40 ----------(4)

Now ,

Adding eq-(2) and (3) , we get ;

=> X + Y + Z + 3X - Y + 2Z = 10 + 13

=> 4X + 3Z = 23 ----------(5)

Now ,

Subtracting eq-(4) from 2×(5) , we get ;

=> 2×(4X + 3Z) - (5X + 6Z) = 2×23 - 40

=> 8X + 6Z - 5X - 6Z = 46 - 40

=> 3X = 6

=> X = 6/3

=> X = 2

=> 1/x = 2

=> x = 1/2

Now ,

Putting X = 2 in eq-(5) , we get ;

=> 4X + 3Z = 23

=> 4×(1/2) + 3Z = 23

=> 2 + 3Z = 23

=> 3Z = 23 - 2

=> 3Z = 21

=> Z = 21/3

=> Z = 7

=> 1/z = 7

=> z = 1/7

Now ,

Putting X = 2 , Z = 7 in eq-(2) , we get ;

=> X + Y + Z = 10

=> 2 + Y + 7 = 10

=> Y + 9 = 10

=> Y = 10 - 9

=> Y = 1

=> 1/y = 1

=> y = 1

Hence ,

x = 1/2 , y = 1 , z = 1/7