1. ## Inequality

Hey guys I'm confused whether I doing these right, I used 2 different methods to solve this problem, and they seem to be giving different answers

$\text{solve} \dfrac{1}{x-3}> 4$

1st method:

$\dfrac{1}{x-3}> 4$

$\dfrac{1}{(x-3)^2}>16$

$1>16(x-3)^2$

$1>16x^2-96x+144$

$16x^2-96x+143<0$

$x=\dfrac{96\pm\sqrt{96^2-4(16)(143)}}{32}$

$\dfrac{88}{32}

2nd method

$\dfrac{1}{x-3}> 4$

$\dfrac{1}{x-3}-4>0$

$\text{multiply both sides by}\hspace{5mm}(x-3)^2$

$(x-3)-4(x-3)^2>0$

$(x-3)(1-4(x-3))>0$

$(x-3)(13-4x)>0$

$3

they give different answers...? Pls let me know if i've done anything wrong

Thanks

2. Whenever you square an equation or inequation to aid in solving it, you are likely to bring in extraneous solutions.

What I would do is invert both sides, noting that inverting both sides of an inequation changes the direction of the inequality sign. The rest should be obvious.

3. First we need to get a common denominator on both sides

$\displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}$

simplify to get

$\displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}$

So the two critical numbers(where the inequality can chage) are $\displaystyle x=3,\quad x=\frac{13}{4}$

If you test some points you will get

$\displaystyle \left(3,\frac{13}{4}\right)$

4. Originally Posted by TheEmptySet
First we need to get a common denominator on both sides

$\displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}$

simplify to get

$\displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}$

So the two critical numbers(where the inequality can chage) are $\displaystyle x=3,quad x=\frac{13}{4}$

If you test some points you will get

$\displaystyle (-\infty,3) \cup \left(\frac{13}{4},\infty \right)$
Sorry but for any $\displaystyle x$ in $\displaystyle (-\infty, 3)$, $\displaystyle \frac{1}{x - 3}$ gives something negative, which is clearly NOT greater than $\displaystyle 4$...

5. The way I would have done it is this: $\frac{1}{x-3}> 4$ is the same as $0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $(-\infty, 3)$, $(3, 13/4)$, and $(13/4, \infty)$. If x= 0, which is less than 3, $\frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\frac{13}{4}= 3\frac{1}{4}$, $\frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.

Therefore, the inequality is true on the interval $\left(3, \frac{13}{4}\right)$.

6. Originally Posted by Prove It
Sorry but for any $\displaystyle x$ in $\displaystyle (-\infty, 3)$, $\displaystyle \frac{1}{x - 3}$ gives something negative, which is clearly NOT greater than $\displaystyle 4$...
You are correct I cant add this morning! thanks

7. "If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.

The second method is wrong. You don't know if $\frac{(x-3)^2}{x-3}=x-3$ or $3-x$.

First method is wrong too. You can square only after you say that x must be bigger then 3.

8. Originally Posted by HallsofIvy
The way I would have done it is this: $\frac{1}{x-3}> 4$ is the same as $0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $(-\infty, 3)$, $(3, 13/4)$, and $(13/4, \infty)$. If x= 0, which is less than 3, $\frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\frac{13}{4}= 3\frac{1}{4}$, $\frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.

Therefore, the inequality is true on the interval $\left(3, \frac{13}{4}\right)$.
Actually $\displaystyle \frac{1}{x - 3} > 4$ is the same as $\displaystyle 0 > 4 - \frac{1}{x - 3}$...

9. A way this kind of inequalities can be solved:

$\frac{1}{x-3}>4 \Leftrightarrow 4-\frac{1}{x-3}<0 \Leftrightarrow \frac{4x-13}{x-3}<0$

You want that fraction be negativ, so $x \in \left ( 3, \frac{13}{4} \right )$.

10. ## Re: Inequality

Originally Posted by veileen
The second method is wrong. You don't know if $\frac{(x-3)^2}{x-3}=x-3$ or $3-x$.
sorry for bringing up an old thread, but I thought about this inequality thing again, looking over what you wrote isn't $\frac{(x-3)^2}{x-3}=x-3$ always?, isn't this true for all real numbers x, for example sub in x=5 then the answer would be 2, which is 5-3, if subbing in lets say a negative number -1, then it would be 16/-4=-1-3 which is true again, the same is true for x=0 or a positive number smaller than 3, eg 1

11. ## Re: Inequality

The statement:
$\frac{(x-3)^2}{x-3}=3-x$ is not true because, that would mean $(3-x)\cdot(x-3)=(x-3)^2 \Leftrightarrow -x^2+6x-9=x^2-6x+9$ which is not true for all $x$.

12. ## Re: Inequality

Originally Posted by veileen
"If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.
There are a lot of things that anybody can do- that doesn't mean it is wrong.

The only places an inequality can change form ">" to "<" is where the function is 0 or is not continuous. For fractions that is where the numerator is 0 or where the denominator is 0. Once you have found those points, checking one point in each interval tells you which. What TheEmptySet said was perfectly valid.