solve each system using the substitution method

[ginny1029]

x = y + 1
x + 2y = 4

since I already have the x as the x = y + 1 i substitution

x + 2y = 4
(y + 1) + 2y = 4
y + 1 + 2y = 4
3y + 1 = 4
3y = 4 - 1
3y = 3
3y/3 = 3/3
y = 1

then go back to the first equation
x = y + 1
x = 1 + 1
x = 2

(2,1)

does that look right?

 

[galactus]

Yep