Question Number 8230 by 314159 last updated on 03/Oct/16
$$\mathrm{If}\:\frac{{x}+{y}}{{x}+{y}+{z}}\:=\:\frac{{y}+{z}}{{x}+{y}+{z}}\:=\:\frac{{x}+{z}}{{x}+{y}+{z}}\:={p},\:\mathrm{then}\:\mathrm{which}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{can}\:\mathrm{be}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{p}? \\ $$
Answered by Rasheed Soomro last updated on 03/Oct/16
$$\frac{\mathrm{a}}{\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{C}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{A}+\mathrm{B}+\mathrm{C}} \\ $$$$\:\mathrm{p}=\frac{{x}+{y}}{{x}+{y}+{z}}\:=\:\frac{{y}+{z}}{{x}+{y}+{z}}\:=\:\frac{{x}+{z}}{{x}+{y}+{z}} \\ $$$$\:\:\:\:\:\:=\frac{\left(\mathrm{x}+\mathrm{y}\right)+\left(\mathrm{y}+\mathrm{z}\right)+\left(\mathrm{x}+\mathrm{z}\right)}{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)+\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)+\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{2x}+\mathrm{2y}+\mathrm{2z}}{\mathrm{3}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}=\frac{\mathrm{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}{\mathrm{3}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$