I need help with this math question (pic induced)

Answer:
A
Step-by-step explanation:
Set the equation equal to bx.
[tex](3x + 3)(ax - 2) - {x}^{2} + 6 = bx[/tex]
[tex]3a {x}^{2} - 6x + 3ax - 6 - {x}^{2} + 6 = bx[/tex]
Cancel out the constants
[tex]3 {ax}^{2} - 6x + 3ax - x {}^{2} = bx[/tex]
In order to cancel out the quadratics, we must solve for a.
[tex]3 {ax}^{2} - {x}^{2} = 0[/tex]
[tex]3 {ax}^{2} = {x}^{2} [/tex]
[tex]3a = 1[/tex]
[tex]a = \frac{1}{3} [/tex]
So plug in 1/3 for a.
[tex]3( \frac{1}{3} ) {x}^{2} - 6x + 3( \frac{1}{3} )x - {x}^{2} = bx[/tex]
[tex] - 6x + x = bx[/tex]
[tex] - 5x = bx[/tex]
[tex]b = - 5[/tex]