Hi @gregmat,

The following example in the prepswift algebra video is incorrect.

I think x can be zero.

Assuming x = 0, the equation can be solved like

(2*0 + 3) / (4 - 5*0) = 3 / 4

Hi @gregmat,

The following example in the prepswift algebra video is incorrect.

I think x can be zero.

Assuming x = 0, the equation can be solved like

(2*0 + 3) / (4 - 5*0) = 3 / 4

The original equation is the one on the top left. We know that the denominator cannot equal zero.

12x-15x^2 cannot equal 0, so 3x(4-5x) cannot equal 0. The two values that x cannot be are 0, and 4/5.

You can also try plugging zero into the original equation. (6*0*0+9*0) / (12*0-15*0*0) = 0/0, which is undefined.

Correct. But suppose we just had

2x + 3 / 4 - 5x

as the expression, and it wasn’t the result of simplifying a complex expression. Then x can be zero, right?

Yes.