Question

Help!!! What is the difference ? (x - 8)/(3x) - (3x)/(x ^ 2)

Answer 1

Answer:

The difference of $\frac{x-8}{3x}-\frac{3x}{x^2}$  is $\mathbf{\frac{x-17}{3x}}$

Step-by-step explanation:

We need to find the difference: $\frac{x-8}{3x}-\frac{3x}{x^2}$

Simplifying to find the difference:

$\frac{x-8}{3x}-\frac{3x}{x^2}$

x is common in both 3x and x^2 so, we can cancel it out and we get:

$=\frac{x-8}{3x}-\frac{3x}{x(x)}\\=\frac{x-8}{3x}-\frac{3}{x}$

Now, taking LCM of 3x and x we get 3x, We will multiply 3x with both terms and then solve we get:

$=\frac{x-8-3(3)}{3x}\\=\frac{x-8-9}{3x}\\=\frac{x-17}{3x}$

It cannot be further simplified, so we get $\frac{x-17}{3x}$ as answer.

So, The difference of $\frac{x-8}{3x}-\frac{3x}{x^2}$  is $\mathbf{\frac{x-17}{3x}}$

Answer 2

Answer:

Given that you're providing only a single term, asking for a difference doesn't make a lot of sense.  I assume you're asking to have the expression simplified?

If so, then it can be reduced to (x - 17) / 3x

Step-by-step explanation:

You can simplify by multiplying both terms each by a different ratio that equals one, but also gives the terms common denominator.  Multiplying the first one by x²/x² and the second by 3x/3x will do that for you.

After that you can group the terms into one fraction and simplify:

$\frac{x - 8}{3x} - \frac{3x}{x^2} \\\\= \frac{x - 8}{3x}\times\frac{x^2}{x^2} - \frac{3x}{x^2} \times\frac{3x}{3x} \\\\= \frac{x^3 - 8x^2}{3x^3} - \frac{9x^2}{3x^3} \\\\= \frac{x^3 - 8x^2 - 9x^2}{3x^3} \\\\= \frac{x - 8 - 9}{3x}\\\\= \frac{x - 17}{3x}$