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.
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The inequation
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Guest

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)
Guest

Re: Calculus

Another way: after we have $$\frac{4x+ 34}{x}< 0$$, obviously x cannot be 0 and remember that multiplying both sides of an inequality by a negative number reverses the direction of the inequality while multiplying both sides of an inequality by a positive number does not change it.

So consider two cases:
1) x> 0. Then, multiplying both sides by x, a positive number, 4x+ 34< 0. Subtracting 34 from both sides results in 4x< -34 so x< -17/2. All those values are negative so x> 0 is not true. There are no positive values of x that make this inequality true.

2) x< 0. Then, multiplying both sides by x, a negative number, 4x+ 34> 0. Subtracting 34 from both sides results in 4x> -34 so x> -17/2. Since x must also be negative, this inequality is true for -17/2< x< 0.

("Guest" (who posts an awful lot!) lost the sign going from "4x+ 34= 0" to "x= 17/2".)

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