Just to add to the discussion, here is the solution using the substitution method.

Solve x+y-2=0 for x. x=-y+2

Substitute -y+2 for the x in the equation 2x-3y+1=0.

2(-y+2)-3y+1=0

Distribute the 2: -2y+4-3y+1=0

-5y+5=0

-5y=-5

y=1

Substitute the value of 1 into either equation for the y value. I will use x+y-2=0.

x+1-2=0

x-1=0

x=1

The solution is the point (1,1).

The system can be solved either using elimination method, or substitution method.

We'll choose elimination method to remove y variable. For this reason we'll multiply the 2nd equation by 3:

3x + 3y - 6 = 0

We'll add this equation to the 1st equation of the system:

3x + 3y - 6 + 2x - 3y + 1 = 0

We'll eliminate variable y:

5x - 5 = 0

We'll isolate 5x to the left side, adding 5 both sides:

5x = 5

x = 1

We'll replace x by 1 within the second equation, to determine y:

1 + y - 2 = 0

y = 2 - 1

y = 1

**The solution of the system of independent equations is (1 ; 1).**