Question

With steps , please.

Answer 1

`\bold{\text{Answer:}\quad x=(1)/(2),\quad y=1,\quad z=(1)/(3)}`

Explanation:

`\text{Equation 1:}\quad (x)/(x+y)=(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=(x)/(3x-1)`

`\text{Equation 2:}\quad (y)/(y+z)=(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=(z)/(4z-1)`

`\text{Equation 3:}\quad (z)/(z+x)=(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=(x)/(5x-1)`

Set Equation 1 equal to Equation 2 and substitute z per Equation 3

`(x)/(3x-1)=(z)/(4z-1)\\\\\\x(4z-1)=z(3x-1)\\\\4xz-x=3xz-z\\\\4x\bigg((x)/(5x-1)\bigg)-x=3x\bigg((x)/(5x-1)\bigg)-(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=(1)/(2)`

Solve for y when x = 0:

`\text{Equation 1:}\quad y=(0)/(3(0)-1)\quad \rightarrow \quad y=0`

Notice that x + y is in the denominator and denominator cannot equal zero so x = 0 is an invalid solution.

`\text{Solve for y when}\ x=(1)/(2):\\\\\text{Equation 1:}\quad y=((1)/(2))/(3((1)/(2))-1)\quad \rightarrow \quad y=1`

`\text{Solve for z when x = (1)/(2)}:\\\text{Equation 3:}\quad z=((1)/(2))/(5((1)/(2))-1)\quad \rightarrow \quad z=(1)/(3)`
`\text{Solve for z when}\ x=(1)/(2):\\\\\text{Equation 3:}\quad z=((1)/(2))/(5((1)/(2))-1)\quad \rightarrow \quad z=(1)/(3)`