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