Solve With Substitution

(6, 3)

(1, 2)

(3, 6)

(2, 1)

No solution

(1, 2)

(3, 0)

(2, 1)

(1, -2)

(2, -2)

(1, 2)

Why have you done this?

(5, -1)

(-5, 1)

(-5,-1)

(1,-5)

(3,-9)

(-3,-9)

(-3,9)

(0,1)

y=3

y=1

y= - 2

y=4

(6, 6)

(6, 9)

(9, 6)

(9, 9)

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

(-34, -17)

(8, 4)

(10, 20)

No solution

Infinite solutions

(-6, 1)

(-12, -12)

(1, -6)

No solution

Infinite solutions