Question

I need help with this math question (pic induced)

Answer 1

Answer:

A

Step-by-step explanation:

Set the equation equal to bx.

$(3x + 3)(ax - 2) - {x}^{2} + 6 = bx$

$3a {x}^{2} - 6x + 3ax - 6 - {x}^{2} + 6 = bx$

Cancel out the constants

$3 {ax}^{2} - 6x + 3ax - x {}^{2} = bx$

In order to cancel out the quadratics, we must solve for a.

$3 {ax}^{2} - {x}^{2} = 0$

$3 {ax}^{2} = {x}^{2}$

$3a = 1$

$a = \frac{1}{3}$

So plug in 1/3 for a.

$3( \frac{1}{3} ) {x}^{2} - 6x + 3( \frac{1}{3} )x - {x}^{2} = bx$

$- 6x + x = bx$

$- 5x = bx$

$b = - 5$