`2x - y = 0, x - y = 7` Solve the system of linear equations and check any solutions algebraically.

EQ1: 2x-y=0

EQ2: x-y=7

To solve this system of equations, let's apply substitution method. Let's isolate the x in the second equation.

`x - y=7`

`x=7+y`

Then, plug-in this to the first equation.

`2x - y=0`

`2(7+y)-y = 0`

And solve for y.

`14+2y-y=0`

`14+y=0`

`y=-14`

Now that the value of y is known, solve for x. Let's plug-in y=-14 to the second equation.

`x -y=7`

`x-(-14)=7`

`x+14=7`

`x=7-14`

`x=-7`

To check, plug-in x=-7 and y=-14 to one of the original equations. Let's use the first equation.

`2x-y=0`

`2(-7) - (-14)=0`

`-14 + 14=0`

`0=0 `   (True)

Therefore, the solution is (-7,-14).