Help needed solving this system of equations

I've tried solving this equation, but keep ending up with no solution. Can any of you try to figure this one out and explain to me how you solved it? Thanks in advance.

x^2-xy+3y^2=5
x-y=2

Solve the systems of equations algebraically.

Take the second equation to get y = x - 2, then substitute back in the original equation.

x^2-xy+3y^2=5
x^2 - x(x-2) + 3(x-2)^2 = 5
x^2 - x^2 + 2x + 3x^2 - 12x + 12 = 5
3x^2 - 10x + 7 = 0
(3x - 7)(x - 1) = 0
x = 1 or 7/3

Using y = x - 2, we see that the solutions are (1,-1) and (7/3, 1/3)

x^2-xy+3y^2=5
x^2-x(y=x-2)+3(x-2)^2=5
-^2 -^2
x-x(y=x-2)+3(x-2)=5
+2 +2
x-x(y)=3(x)=5
y+3(x)=5
-3 -3
y+x=2
(x=2)


2^2-2y+3y^2=5
-^2 -^2
2-2y+3y=5
+2y +2y
2-5y=5
-2 -2
-5y=3
+5 +5
y=8


y=8 x=2
im really not sure if this is right or not. its been a long time since i did this.