[tex]\bold{\text{Answer:}\quad x=\dfrac{1}{2},\quad y=1,\quad z=\dfrac{1}{3}}[/tex]
Step-by-step explanation:
[tex]\text{Equation 1:}\quad \dfrac{x}{x+y}=\dfrac{1}{3y}\\\\\\.\qquad \qquad \qquad 3xy=x+y\\\\.\qquad \qquad \qquad 3xy-y=x\\\\.\qquad \qquad \qquad y(3x-1)=x\\\\.\qquad \qquad \qquad y=\dfrac{x}{3x-1}[/tex]
[tex]\text{Equation 2:}\quad \dfrac{y}{y+z}=\dfrac{1}{4z}\\\\\\.\qquad \qquad \qquad 4yz=y+z\\\\.\qquad \qquad \qquad 4yz-y=z\\\\.\qquad \qquad \qquad y(4z-1)=z\\\\.\qquad \qquad \qquad y=\dfrac{z}{4z-1}[/tex]
[tex]\text{Equation 3:}\quad \dfrac{z}{z+x}=\dfrac{1}{5x}\\\\\\.\qquad \qquad \qquad 5xz=z+x\\\\.\qquad \qquad \qquad 5xz-z=x\\\\.\qquad \qquad \qquad z(5x-1)=x\\\\.\qquad \qquad \qquad z=\dfrac{x}{5x-1}[/tex]
Set Equation 1 equal to Equation 2 and substitute z per Equation 3
[tex]\dfrac{x}{3x-1}=\dfrac{z}{4z-1}\\\\\\x(4z-1)=z(3x-1)\\\\4xz-x=3xz-z\\\\4x\bigg(\dfrac{x}{5x-1}\bigg)-x=3x\bigg(\dfrac{x}{5x-1}\bigg)-\dfrac{x}{5x-1}\\\\\\4x^2-x(5x-1)=3x^2-x\\\\4x^2-5x^2+x=3x^2-x\\\\0=4x^2-2x\\\\0=2x(2x-1)\\\\0=2x\qquad\qquad 0=2x-1\\\\x=0\qquad \qquad x=\dfrac{1}{2}[/tex]
Solve for y when x = 0:
[tex]\text{Equation 1:}\quad y=\dfrac{0}{3(0)-1}\quad \rightarrow \quad y=0[/tex]
Notice that x + y is in the denominator and denominator cannot equal zero so x = 0 is an invalid solution.
[tex]\text{Solve for y when}\ x=\dfrac{1}{2}:\\\\\text{Equation 1:}\quad y=\dfrac{\frac{1}{2}}{3(\frac{1}{2})-1}\quad \rightarrow \quad y=1[/tex]
[tex]\text{Solve for z when x = \dfrac{1}{2}}:\\\text{Equation 3:}\quad z=\dfrac{\frac{1}{2}}{5(\frac{1}{2})-1}\quad \rightarrow \quad z=\dfrac{1}{3}[/tex] [tex]\text{Solve for z when}\ x=\dfrac{1}{2}:\\\\\text{Equation 3:}\quad z=\dfrac{\frac{1}{2}}{5(\frac{1}{2})-1}\quad \rightarrow \quad z=\dfrac{1}{3}[/tex]
