Question

I'm so confused, I've never encountered a problem like this; Do I flip it, do I solve it like that?

https://prnt.sc/fb86wl

Answer 1

So you were given and asked to solve:
\(\dfrac{6}{x - 6} = \dfrac{x}{x - 6} - \dfrac{6}{2}\)

What do you think we might be able to do first? Are one of the fractions reducable?

Answer 2

6/2 can be reduced to just 3

Answer 3

Precisely, so reducing that fraction to three leaves us with:
\(\dfrac{6}{x - 6} = \dfrac{x}{x - 6} - 3\)

Do you remember what we can do with two fractions that have the same denominator?

Answer 4

you can cancel them out?

Answer 5

You can either add or subtract two fractions that have the same denominator. This is a hint on what to do next.

Answer 6

okay so if we're using -6 we can do 6/-12 and we can just add 6/-12 and -6/-12?

Answer 7

From where do you get \(\dfrac{-6}{-12}\)? We don't know what \(x\) is. We're trying to solve for \(x\). As I expressed above, we see that the two remaining fractions have the same denominator so we must add or subtract them in order to continue solve for \(x\). Which operation should we perform, addition or subtraction?

Answer 8

It gives us x in the answers it's either -6 or 6, I just fill in the blank which is easier for me

Answer 9

Well, actually, we should finish what we started. The next step is to add 3 to both sides, then subtract \(\dfrac{6}{x - 6}\) from both sides to get:

\(3 = \dfrac{x}{x - 6} - \dfrac{6}{x - 6}\)

Answer 10

The next step is to combine the fractions on the right hand side. Doing so yields:
\(3 = \dfrac{x - 6}{x - 6}\).

What do you think we can do from here?

Answer 11

Put 3 in for the x?

Answer 12

Actually, anything over itself is 1 so \(\dfrac{x - 6}{x - 6}\) reduces to 1 which means:

\(3 \ne 1\) and this implies that the solution is extraneous. So we've eliminated two of the options. We know we can't use \(x = 6\) because that would result in a zero denominator which is undefined so the option you chose \(x = -6\) is extraneous was the correct choice