Solve this inequality: 5/6x+2/3x>3/4.

So I simplified the left side and got 3/2x, then moved the equation to one side and got 3/2x-3/4>0. I got x>1/2, but the answer is wrong.

Can somebody help me?

Thanks in advance.

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- Oct 23rd 2011, 12:38 PMDragon08Solving Rational Inequalities
Solve this inequality: 5/6x+2/3x>3/4.

So I simplified the left side and got 3/2x, then moved the equation to one side and got 3/2x-3/4>0. I got x>1/2, but the answer is wrong.

Can somebody help me?

Thanks in advance. - Oct 23rd 2011, 12:42 PMSammySRe: Solving Rational Inequalities
Lets see your work. How did you get x>1/2 , from 3/2x-3/4>0 ?

- Oct 23rd 2011, 12:44 PMDragon08Re: Solving Rational Inequalities
3/2x>3/4. Then, I divided both sides by 3/2.

- Oct 23rd 2011, 12:54 PMSammySRe: Solving Rational Inequalities
That gives (1/x) > (1/2) .

- Oct 23rd 2011, 12:55 PMDragon08Re: Solving Rational Inequalities
How did you get that?

- Oct 23rd 2011, 01:06 PMSammySRe: Solving Rational Inequalities
Using algebra ... and assuming what you meant was 3/(2x) > 3/4 .

Was the original problem: Solve 5/(6x) + 2/(3x) > 3/4 ? - Oct 23rd 2011, 02:07 PMDragon08Re: Solving Rational Inequalities
Yes

- Oct 23rd 2011, 02:33 PMSammySRe: Solving Rational Inequalities
Starting with $\displaystyle \frac{3}{2x}>\frac{3}{4}$

Multiply by 2/3 $\displaystyle \frac{2}{3}\cdot\frac{3}{2x}>\frac{2}{3}\cdot\frac {3}{4}$

This gives $\displaystyle \frac{1}{x}>\frac{1}{2}$

Then a little known rule says that if $\displaystyle \frac{a}{b}>\frac{c}{d}$ then $\displaystyle \frac{b}{a}<\frac{d}{c}$ - Oct 23rd 2011, 02:53 PMDragon08Re: Solving Rational Inequalities
Okay, I get it. Thank you

- Nov 4th 2011, 06:21 AMHallsofIvyRe: Solving Rational Inequalities
A large part of your problem is that what you give as the question itself is ambiguous- use parentheses!

If the problem was (5/6)x+(2/3)x>3/4, then x> 1/2 is exactly right. However, it appears that your problem is**really**5/(6x)+ 2/(3x)> 3/4. If that is what the inequality is, then, as SammyS said, you arrive at 1/x> 1/2 so that 0< x< 2. (If 1/x> 1/2 then x must be positive.)