#(x+y) prop z,(y+z) prop x# then prove that #(z+x) prop y # ?thanks

Question

6296 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-09T05:15:58

Given

#x+ypropz#

#=>x+y=mz.......[1]#, where m = proportionality constant

#=>(x+y)/z=m#

#=>(x+y+z)/z=m+1 ....[2]#

Again

#y+zpropx#

#=>y+z=nx........[3]#, where n = proportionality constant

#=>(y+z)/x=n#

#=>(x+y+z)/x=n+1 ......[4]#

Dividing [2] by [4]

#x/z=(m+1)/(n+1)=k(say)#

#=>x=kz......[5]#

By [1] and [5] we get

#kz+y=mz#

#=>y=(m-k)z#

#=>y/z=(m-k)......[6]#

Dividing [2] by [6] we get

#(x+y+z)/y=(m+1)/(m-k)=c " another constant"#

#=>(x+y+z)/y-1=c -1#

#=>(x+z)/y=c -1="constant"#

Hence

#z+xpropy#

Proved