Which expression is equivalent to $\left(\frac{2}{3}-\frac{2}{x}\right) \div \frac{x-3}{x}$? (from Khan academy)

Question

I've been trying to understand this problem for hours but not getting it. HELP!!!

The correct answer is $\frac{2}{3}$, but I don't know why this is the correct answer.

Thank you in advance for your help!

Answer 1

Sharing my viewes, $$\left(\frac{2}{3}-\frac{2}{x}\right) \div \frac{x-3}{x}$$ $$=\left(\frac{2}{3}-\frac{2}{x}\right) \times \frac{x}{x-3}$$ $$=2\left(\frac{x-3}{3x}\right) \times \frac{x}{x-3}$$ $$=\frac{2}{3}$$

Answer 2

We start with $$\left(\frac{2}{3}-\frac{2}{x}\right) \div \frac{x-3}{x}.$$ As usual, we evaluate what's in the brackets first. Particularly, if we can change this into a fraction, then we'll be dividing fractions, which we can do without too much trouble. So, we wish to simplify $$\frac{2}{3}-\frac{2}{x}$$ into a single fraction. Note the different denominator. We'll need a common denominator. Since $3$ and $x$ may not share any common factors, we'll just use their product, $3x$. We multiply the $\frac{2}{3}$ top and bottom by $x$, and the $\frac{2}{x}$ top and bottom by $3$. This gives us $$\frac{2}{3}-\frac{2}{x} = \frac{2x}{3x} - \frac{6}{3x}.$$ Now that the denominators are the same, we subtract as usual: subtract the numerators, and keep the common denominator, so $$\frac{2}{3}-\frac{2}{x} = \frac{2x - 6}{3x}.$$ This gives us $$\left(\frac{2}{3}-\frac{2}{x}\right) \div \frac{x-3}{x} = \frac{2x - 6}{3x} \div \frac{x-3}{x}.$$ To divide fractions, we invert and multiply. Thus, $$\frac{2x - 6}{3x} \div \frac{x-3}{x} = \frac{2x - 6}{3x} \times \frac{x}{x-3}.$$ Multiplying is as simple as multiplying numerators and denominators: $$= \frac{(2x - 6)x}{3x(x - 3)}.$$ Finally, we can perform some factoring and cancelling. In particular, $2x - 6$ has a common factor of $2$, and is equal to $2(x - 3)$. So, $$= \frac{2(x - 3)x}{3x(x - 3)}.$$ Finally, cancelling common factors of $x$ and $x - 3$, we get $$= \frac{2}{3}.$$ Hope that helps.