Question

(X-3) (2x-3)= (x-3) (x+1) solve for x

Answer 1

$(x - 3)(2x - 3) = (x - 3)(x + 1)$

Do the multiplication both sides,

$2 {x}^{2} - 9x + 9 = {x}^{2} - 2x - 3$

Arrange the terms to LHS according to their variables,

$2 {x}^{2} - {x}^{2} - 9x + 2x + 9 - 3 = 0 \\ {x}^{2} - 7x + 6 = 0$


Factorise this polynomial

${x}^{2} - 6x - 6x + 6 = 0 \\ x(x - 6) - (x - 6) = 0 \\ (x - 1)(x - 6) = 0$


So what we notice here is when you're multiplying two numbers you're getting 0 as answer.

So one of them must be 0.

If (x-1) is 0
then x = 1

similarly,

x-6=0

then, x= 6.


So the possible values of x are 1 and 6.

Answer 2

$2 {x}^{2} - 3x - 6x + 9 = {x}^{2} + x - 3x - 3 \\ {x}^{2} - 7x + 12 = 0 \\ (x - 3)(x - 4) = 0 \\ x = 3 \: or \: x = 4$