SOLUTION: How do I solve a problem when the variables are apart of the fraction? 2/y + 3/x + 8 = 0 4/z - 2/x + 5 = 0 1/y + 4/z = -5

Algebra ->  Matrices-and-determiminant -> SOLUTION: How do I solve a problem when the variables are apart of the fraction? 2/y + 3/x + 8 = 0 4/z - 2/x + 5 = 0 1/y + 4/z = -5       Log On


   

Question 722069: How do I solve a problem when the variables are apart of the fraction?
2/y + 3/x + 8 = 0
4/z - 2/x + 5 = 0
1/y + 4/z = -5
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let a= 1/x
b=1/y
and c=1/z
2b+3a=-8................(1)
4c-2a=-5.....................(2)
b+4c=-5.......................(3)
4c-2a=-5( multiply by -1)
-4c+2a=5
b+4c=-5
add both we get
2a+b=0
2b+3a=-8 ...from (1)
eliminate b-a=-8
a=8
so b=-16
c=11/4
Reciprocal of a,b.c are x,y,z
x=1/8, y=-1/16, z= 4/11