solving inequalities

**puresoul** · December 2nd 2012, 10:13 PM

x/(x-2) <5

For solving this question
I started by finding the point where x/(x-2) =5
and got 5/2
and I know that x cannot be 2
I found that when x is between 2 and 5/2
x/(x-2) > 5

so x has to be<2 and >5/2

Is there another way of solving it??

I studied the table method which uses the critical values but didn't know how to use it here..
Can someone show me how to use it here?
Or is it not possible for this question??

**earboth** · December 3rd 2012, 06:40 AM

Originally Posted by puresoul

x/(x-2) <5

For solving this question
I started by finding the point where x/(x-2) =5
and got 5/2
and I know that x cannot be 2
I found that when x is between 2 and 5/2
x/(x-2) > 5

so x has to be<2 and >5/2

Is there another way of solving it??

I studied the table method which uses the critical values but didn't know how to use it here..
Can someone show me how to use it here?
Or is it not possible for this question??

This is my way:

$\frac x{x-2} > 5$

$\frac x{x-2} - 5 > 0$

$\frac x{x-2} - \frac{5(x-2)}{x-2} > 0$

$\frac{x - 5x+10}{x-2} > 0$

$\frac{ - 4x+10}{x-2} > 0$

A quotient is greater than zero, that is positive, if both numerator and denominator are positive, that means greater than zero, or: if both numerator and denominator are negative, that means smaller than zero

A) Both positive:

$-4x+10>0~\wedge~x-2>0~\implies~x<\frac52~\wedge~x>2$

That means: $2<x<\frac52$

B) Both negative:

$-4x+10<0~\wedge~x-2<0~\implies~x>\frac52~\wedge~x<2$

That means: $x \in \lbrace \rbrace$

Now collect both sets of solutions.

**puresoul** · December 4th 2012, 01:05 AM

Originally Posted by earboth

This is my way:

$\frac x{x-2} > 5$

$\frac x{x-2} - 5 > 0$

$\frac x{x-2} - \frac{5(x-2)}{x-2} > 0$

$\frac{x - 5x+10}{x-2} > 0$

$\frac{ - 4x+10}{x-2} > 0$

A quotient is greater than zero, that is positive, if both numerator and denominator are positive, that means greater than zero, or: if both numerator and denominator are negative, that means smaller than zero

A) Both positive:

$-4x+10>0~\wedge~x-2>0~\implies~x<\frac52~\wedge~x>2$

That means: $2<x<\frac52$

B) Both negative:

$-4x+10<0~\wedge~x-2<0~\implies~x>\frac52~\wedge~x<2$

That means: $x \in \lbrace \rbrace$

Now collect both sets of solutions.

Sorry but what do you mean by collecting both sets of solutions??
Check which one satisfies the inequality given ??

**MarkFL** · December 4th 2012, 01:35 AM

My way is similar to that of earboth:

$\frac{x}{x-2}>5$

$\frac{x}{x-2}-5>0$

$\frac{x}{x-2}-\frac{5(x-2)}{x-2}>0$

$\frac{x-5x+10}{x-2}>0$

$\frac{10-4x}{x-2}>0$

$\frac{2(5-2x)}{x-2}>0$

Equating the numerator and denominator to zero, we find two critical numbers:

$x=2,\,\frac{5}{2}$

Dividing the real number line at these critical value gives us 3 intervals. Since neither of the critical numbers is a root of even multiplcity, we know the sign of the expression will alternate across the intervals. We pick a test number from the first (leftmost) interval:

$(-\infty,2)$ test value $x=0$ expression is negative, so our solution will be:

$\left(2,\frac{5}{2} \right)$

We use an open interval since the inequality is strict. If it were weak, we would still have to exclude 2 to avoid division by zero.

**puresoul** · December 4th 2012, 02:13 AM

Originally Posted by MarkFL2

My way is similar to that of earboth:

$\frac{x}{x-2}>5$

$\frac{x}{x-2}-5>0$

$\frac{x}{x-2}-\frac{5(x-2)}{x-2}>0$

$\frac{x-5x+10}{x-2}>0$

$\frac{10-4x}{x-2}>0$

$\frac{2(5-2x)}{x-2}>0$

Equating the numerator and denominator to zero, we find two critical numbers:

$x=2,\,\frac{5}{2}$

Dividing the real number line at these critical value gives us 3 intervals. Since neither of the critical numbers is a root of even multiplcity, we know the sign of the expression will alternate across the intervals. We pick a test number from the first (leftmost) interval:

$(-\infty,2)$ test value $x=0$ expression is negative, so our solution will be:

$\left(2,\frac{5}{2} \right)$

We use an open interval since the inequality is strict. If it were weak, we would still have to exclude 2 to avoid division by zero.

Took me soo long to understand why your answers are opposite to mine!!
and i just realised that you were finding when x/(x-2) >5
and i asked when x/(x-2) <5
><

Anyways thanks for your help, I understood your way

**Prove It** · December 4th 2012, 02:32 AM

Originally Posted by puresoul

x/(x-2) <5

For solving this question
I started by finding the point where x/(x-2) =5
and got 5/2
and I know that x cannot be 2
I found that when x is between 2 and 5/2
x/(x-2) > 5

so x has to be<2 and >5/2

Is there another way of solving it??

I studied the table method which uses the critical values but didn't know how to use it here..
Can someone show me how to use it here?
Or is it not possible for this question??

Another method: First it should be obvious that $\displaystyle \begin{align*} x \neq 2 \end{align*}$ , now

$\displaystyle \begin{align*} \frac{x}{x - 2} &< 5 \\ 1 + \frac{2}{x - 2} &< 5 \\ \frac{2}{x - 2} &< 4 \\ \frac{1}{x - 2} &< 2 \end{align*}$

Case 1: $\displaystyle \begin{align*} x < 2 \end{align*}$

$\displaystyle \begin{align*} \frac{1}{x - 2} &< 2 \\ 1 &> 2(x - 2) \\ \frac{1}{2} &> x - 2 \\ \frac{5}{2} &> x \end{align*}$

Obviously all $\displaystyle \begin{align*} x < 2 \end{align*}$ will satisfy the inequality.

Case 2: $\displaystyle \begin{align*} x > 2 \end{align*}$

$\displaystyle \begin{align*} \frac{x}{x - 2} &< 5 \\ 1 + \frac{2}{x - 2} &< 5 \\ \frac{2}{x - 2} &< 4 \\ \frac{1}{x - 2} &< 2 \\ 1 &< 2(x - 2) \\ \frac{1}{2} &< x - 2 \\ \frac{5}{2} &< x \end{align*}$

So all $\displaystyle \begin{align*} x > \frac{5}{2} \end{align*}$ will also satisfy the inequality.

So the solution to $\displaystyle \begin{align*} \frac{x}{x - 2} < 5 \end{align*}$ is $\displaystyle \begin{align*} x \in (-\infty, 2) \cup \left( \frac{5}{2}, \infty \right) \end{align*}$ .

**puresoul** · December 4th 2012, 02:53 AM

Originally Posted by Prove It

Another method: First it should be obvious that $\displaystyle \begin{align*} x \neq 2 \end{align*}$ , now

$\displaystyle \begin{align*} \frac{x}{x - 2} &< 5 \\ 1 + \frac{2}{x - 2} &< 5 \\ \frac{2}{x - 2} &< 4 \\ \frac{1}{x - 2} &< 2 \end{align*}$

Case 1: $\displaystyle \begin{align*} x < 2 \end{align*}$

$\displaystyle \begin{align*} \frac{1}{x - 2} &< 2 \\ 1 &> 2(x - 2) \\ \frac{1}{2} &> x - 2 \\ \frac{5}{2} &> x \end{align*}$

Obviously all $\displaystyle \begin{align*} x < 2 \end{align*}$ will satisfy the inequality.

Case 2: $\displaystyle \begin{align*} x > 2 \end{align*}$

$\displaystyle \begin{align*} \frac{x}{x - 2} &< 5 \\ 1 + \frac{2}{x - 2} &< 5 \\ \frac{2}{x - 2} &< 4 \\ \frac{1}{x - 2} &< 2 \\ 1 &< 2(x - 2) \\ \frac{1}{2} &< x - 2 \\ \frac{5}{2} &< x \end{align*}$

So all $\displaystyle \begin{align*} x > \frac{5}{2} \end{align*}$ will also satisfy the inequality.

So the solution to $\displaystyle \begin{align*} \frac{x}{x - 2} < 5 \end{align*}$ is $\displaystyle \begin{align*} x \in (-\infty, 2) \cup \left( \frac{5}{2}, \infty \right) \end{align*}$ .

THANKS prove it

just one thing
how did you eliminate the x on top..
I don't understand what you did in the second step (although I got the later part ^^)

**Prove It** · December 4th 2012, 03:25 AM

Originally Posted by puresoul

THANKS prove it

just one thing
how did you eliminate the x on top..
I don't understand what you did in the second step (although I got the later part ^^)

$\displaystyle \begin{align*} \frac{x}{x - 2} &= \frac{x - 2 + 2}{x - 2} \\ &= \frac{x - 2}{x - 2} + \frac{2}{x - 2} \\ &= 1 + \frac{2}{x - 2} \end{align*}$

**puresoul** · December 4th 2012, 06:06 AM

Originally Posted by Prove It

$\displaystyle \begin{align*} \frac{x}{x - 2} &= \frac{x - 2 + 2}{x - 2} \\ &= \frac{x - 2}{x - 2} + \frac{2}{x - 2} \\ &= 1 + \frac{2}{x - 2} \end{align*}$

Oh!! I get it !!
THANKS!!
^_____^

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