Solving systems of equations by substitution 2x-3y=-1 y=x-1

Y = x - 1so we sub in (x - 1) for y in the other equation

2x - 3y = -1
2x - 3(x - 1) = -1 distribute thru the parenthesis
2x - 3x + 3 = -1 subtract 3 from both sides
2x - 3x = -1 - 3simplify
-x = -4 ...multiply everything by -1 to make x positive
x = 4 

now we sub in 4 for x in either of the original equations to find y

y = x - 1
y = 4 - 1
y = 3

so we have a solution of (4,3)