Algebra problem, possible to answer using only the math tools in algebra

Translated from Swedish:

The square of x is 12 more than x itself.
The cube of x is 9 times x itself.
What # is x ?

So I made these equations but I just couldnt manage to solve an x value that can be plugged back into the equations.

x2 - 12 = x
x3 = 9x

Every answer I got couldn't be plugged back into the equation. Any help appreciated, thanks.
I know the actual answer that works for this equation but I just can't to it using only these equations and the rules of algebra..

x^2-x-12=0
(x-4)(x+3)=0
x=4 or x=-3

x^3-9x=0
x(x^2-9)=0
x(x-3)(x+3)=0
x=0 x=3 x=-3

common solution x=-3

x^2 = x+12

x^2 - x - 12 = 0

x^2 -4x +3x -12 = 0
x(x-4) +3(x-4) = 0

(x-4)(x+3) = 0

=>> x = 4 or x = -3

now substitute in another equation.

x^3 = 9x
4^3 = 9(4)
64 = 36
This one is non equal

x^3 = 9x
(-3)^3 = 9(-3)
-27 = -27
This one is equal

so the answer for x is -3

The square of x is 12 more than x itself.
The cube of x is 9 times x itself.
What # is x ?

x^2 = x + 12
x^2 - x - 12 = 0
(x - 4)(x + 3) = 0
x = Either 4 or - 3

Also, x^3 = 9x
-3 is the value of x.

x^3=9x --> x^2=9 (first divide by x), then : x^2-9=0
(x-3)*(x+3)=0
x1=3
x2=-3

Now, you try if both of them fit:
1) x^2-12=x; 3^2-12=-3 It dosent fit --> 3 != -3
2) (-3)^2-12=-3; -3= -3 --> It fits.

Answer:
x = -3

x2 - 12 = x ......EQN1
x3 = 9x .......EQN2

From eqn2, x^2 = 9
=> x = +/- 3

Substitute x in eqn1
x^2 -12 = x
When x = 3
=> 9 - 12 = 3 (Hence x is not = 3)

When x = -3
9 -12 = -3
=> -3 = -3

Hence x = -3

x^2=x+12
x^2-x-12=0
x^2-4x+3x-12=0
x(x-4)+3(x-4)=0
(x-4)(x+3)=0
x=4 or x=-3

x^3=9x
put x=4
4^3=9*4
64=36(is wrong so x is not 4)

x^3=9x
put x=-3
-3^3=9*-3
-27=-27(is correct so x=-3)