?If \( x y z=4 \) and \( y^{2} z=5 \), what is the value of \( x / y-1 \)

To solve for \( \frac{x}{y} - 1 \), we first need to express \( x \) in terms of \( y \) and \( z \). From the equation \( xyz = 4 \), we can isolate \( x \): \[ x = \frac{4}{yz} \] Now, from the second equation \( y^2 z = 5 \), we can isolate \( z \): \[ z = \frac{5}{y^2} \] Now we will substitute \( z \) back into the expression for \( x \): \[ x = \frac{4}{y \left( \frac{5}{y^2} \right)} = \frac{4y^2}{5} \] Now we can find \( \frac{x}{y} \): \[ \frac{x}{y} = \frac{\frac{4y^2}{5}}{y} = \frac{4y}{5} \] Next, we subtract 1: \[ \frac{x}{y} - 1 = \frac{4y}{5} - 1 = \frac{4y}{5} - \frac{5}{5} = \frac{4y - 5}{5} \] To find the numerical value, we need to determine \( y \). We already have the equation \( y^2 z = 5 \) and we substitute \( z \) from earlier: \[ y^2 \left( \frac{5}{y^2} \right) = 5 \] This equation holds true for any \( y \neq 0 \). To derive more concrete values, we'll substitute \( y = 1 \): Then \( z = \frac{5}{(1)^2} = 5 \) and, \[ x \cdot 1 \cdot 5 = 4 \implies x = \frac{4}{5} \] Now substituting \( y = 1 \) into \( \frac{x}{y} - 1\): \[ \frac{x}{y} - 1 = \frac{\frac{4}{5}}{1} - 1 = \frac{4}{5} - \frac{5}{5} = \frac{4-5}{5} = \frac{-1}{5} \] Thus, the value of \( \frac{x}{y}-1 \) is: \[ \boxed{-\frac{1}{5}} \]