# Calculus

### Calculus

I just do not know how to solve an inequation with variables in two fractions on both sides of the inequation. I apologize for my english, I'm from Brazil and I do not have proficiency in english.
The inequation is attached to the post. If you know how to answer the question or you know a video on YouTube who can help me learn, please send it to me. Thank you.

Attachments
The inequation
20180410_105045-1.jpg (304.51 KiB) Viewed 228 times
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### Re: Calculus

$\frac{36-3x}{x}>\frac{2-7x}{x}$

$\frac{36-3x}{x}-\frac{2-7x}{x}>0$

$\frac{36-3x- 2+7x}{x}>0$

$\frac{4x+34}{x}>0$

First we need to find the root of the numenator
4x + 34 = 0
x = 17/2

So intervals are $(- \infty, 0), (0, 17/2), (17/2, \infty)$
Now let we check a value from the first interval and see if the expression will be positive or negative.
Let we take x = -1
$\frac{4*(-1)+34}{-1} = \frac{30}{-1} < 0$ - in this interval every value we take will give us a negative value

So solutions are only (0, 17/2)

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