You need to write x in terms of y, hence, you may use the top equation, such that:

`x - 5y = 0 => x = 5y`

Replacing 5y for x in the bottom equation, yields:

`5*(5y) = 0 => 25y = 0 => {(25!=0),(y=0):}`

Solving for x the bottom equation, yields:

`5x = 0 => {(5!=0),(x = 0):}`

**Hence, evaluating the solution to the given simultaneous equations, yields `x = 0, y = 0` .**

We notice that the system is a homogeneous system.

We may solve the system using substitution method.

We'll note the equations of the system as if follows:

x-5y=0 (1)

5x=0 (2)

We'll divide by 5 in the equation (2) and we'll get:

x = 0

We'll substitute x by 0 in the first equation:

0 - 5y = 0

-5y = 0

We'll divide by -5 and we'll get:

y = 0

**The solution of the system is represented by the pair {(0,0)}.**