Hi @Bio_J!

To answer the first question, you obtain the answers x=1, y=0 and x=1, y=1 by solving each of the equations and finding the values for x and y.

For the first set of equations:

2x + y = 2x \\
0x + 3y = 2y

We can simplify the first equation to get

2x + y = 2x => y = 2x - 2x => y = 0

Then, to solve for x, we can plug in y = 0 to the original equations to get:

2x + 0 = 2x => x = x \\
0x = 0 => 0 = 0

Choosing some non-zero x that satisfies these equations, we can get x = 1. Therefore we have x = 1, y = 0 as the solution for the first set of equations.

For the second set of equations:

2x + y = 3x \\
0x + 3y = 3y

We can simplify the second equation to get

0x + 3y = 3y => y = y

Then, choosing some non-zero value of y that satisfies this, we can get y = 1.

Next, to solve for x, we can plug in y = 1 to get:

2x + 1 = 3x => 1 = x \\
0x + 3y = 3y => y = y

So we get that x = 1, y = 1 for the second set of equations.

Note that the solutions for x and y will technically be infinite, as any multiple of the values for x and y will solve the equations. For example, for the second set of equations, (1, 1), (2, 2), (10, 10), etc. would all be solutions. Feel free to plug in some other multiples of the solutions to verify this is the case!