Solve Systems of Equations Substitution

The variable is already isolated, so we've already solved the system.

The variable is already isolated for y, so we substitute it in the first equation and solve for x

3x + 3(3x + 5) = 3

Free sha vaca doo

The variable is already isolated for y, so we substitute it in the second equation and solve for x

y = 3(3x + 3y) = 3

(1, -2)

(2, -2)

(1, 2)

Why have you done this?

(5, -1)

(-5, 1)

(-5,-1)

(1,-5)

(5,-2)

(5, 2)

(2,3)

(3,2)

It's the point where both equations equal zero

It's the point where graphs of both equations cross the y-axis

It's the point that solves both equations at the same time

It's the point where graphs of both equations cross the x-axis

(1, 9)

(4, 14)

(0, 1)

(2, 4)

(1, 3)

(-1, -3)

(-1, 3)

(3, 1)

(-2, 5)

(-2, -5)

(2, -5)

(2, 5)

(4, 1)

(1, 4)

(3, 1)

(1, 3)

(4, 1)

(4, -1)

(1, 4)

(-1, 4)