`x - 2y + 5z = 2, 4x - z = 0` Solve the system of linear equations and check any solutions algebraically.

Since the number of equations is smaller than the numbers of variables, the system is indeterminate.

You may write the first equation, such that:

`x + 5z = 2 + 2y`

You need to use a greek letter for y, such that:

`y = alpha`

`x + 5z = 2 + 2alpha => x = 2 + 2alpha - 5z`

You may replace `2 + 2alpha - 5z` in the equation x = z,...

Since the number of equations is smaller than the numbers of variables, the system is indeterminate.

You may write the first equation, such that:

`x + 5z = 2 + 2y`

You need to use a greek letter for y, such that:

`y = alpha`

`x + 5z = 2 + 2alpha => x = 2 + 2alpha - 5z`

You may replace `2 + 2alpha - 5z` in the equation x = z, such that:

`z = 2 + 2alpha - 5z => 6z = 2 + 2alpha`

`z = 1/3 + (alpha)/3`

Hence, evaluating the solutions to the system, yields `x =  1/3 + (alpha)/3, y = alpha, z =  1/3 + (alpha)/3.`