Question

I need help with this math question (pic induced)

Answer 1

A

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 = (1)/(3)`

So plug in 1/3 for a.

`3( (1)/(3) ) {x}^(2) - 6x + 3( (1)/(3) )x - {x}^(2) = bx`

`- 6x + x = bx`

`- 5x = bx`

`b = - 5`