х-у=3у z-2y=y z=5

To solve the system of equations, let's start with the equations you provided:

1. \( x - y = 3y \)
2. \( z - 2y = y \)
3. \( z = 5 \)

First, we'll simplify the equations:

From the first equation, rearranging gives: \[ x - y = 3y \implies x = 4y \]

From the second equation, rearranging gives: \[ z - 2y = y \implies z = 3y \]

Now we can substitute the value of \( z \) from equation 3 into the rearranged second equation: \[ 3y = 5 \] Solving for \( y \): \[ y = \frac{5}{3} \]

Now that we have \( y \), we can substitute it into the equation for \( z \): \[ z = 3y = 3 \cdot \frac{5}{3} = 5 \] This is consistent with the given \( z = 5 \).

Next, we substitute \( y \) into the equation for \( x \): \[ x = 4y = 4 \cdot \frac{5}{3} = \frac{20}{3} \]

So, we have determined the values of the variables:

• \( x = \frac{20}{3} \)
• \( y = \frac{5}{3} \)
• \( z = 5 \)

The solution to the system of equations is: \[ (x, y, z) = \left(\frac{20}{3}, \frac{5}{3}, 5\right) \]