Question

Solve The Following System: Xy/4y-3x =20 Xz/2x-3z =15 Zy/4y-5z =12

Answer 1

Given :

Xy/4y-3x =20

Xz/2x-3z =15

Zy/4y-5z =12

To Find : Solve for x , y and z

Solution:

xy/(4y-3x) = 20

=> (4y - 3x) /xy = 1/20

=> 4/x  - 3/y  = 1/20

xz/(2x-3z) =15

=> (2x - 3z)/xz  = 1/15

=> 2/z  - 3/x  = 1/15

zy/(4y-5z) =12

=> (4y - 5z)/zy = 1/12

=> 4/z - 5/y  = 1/12

4/x  - 3/y  = 1/20    Eq1

2/z  - 3/x  = 1/15    E2

4/z - 5/y  = 1/12       Eq3

2 * Eq1  - Eq3

=> -6/x  + 5/y  = 2/15  -  1/12

=> 5/y - 6/x  = 1/20

5/y - 6/x  =  4/x  - 3/y

=> 8/y  = 10/x      => 5y  = 4x

16/y  - 15/y  = 1/4   =>  y = 4

x = 5

y = 4

z = 3

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