Solving systems of equations by substitution

2x-3y=-1

y=x-1

Y = x - 1 ... so 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 - 3 ... simplify

-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)