In equalities

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• Jan 31st 2011, 04:30 PM
jam2011
In equalities
Book: Calculus with analytic geometry ( fourth edition )by protter and protter.

Problems(page 6 )

x/3-x < 2
= x< 2(3-x) by distribution
= x < 6-2x by adding 2x both side
= 3x < 6 by dividing 3 both side

please check my work if i done correctly. I'm little confuse when can i use cases in inequalities. Thank you in advance
• Jan 31st 2011, 04:34 PM
rtblue
You can't multiply by the denominator when solving an inequality, that is, when the denominator has a variable in it. This is because you do not know whether or not the variable will be negative (if it is, you have to switch the inequality sign).

This is what you would have to do:

$\displaystyle \displaystyle \frac{x}{3-x}<2$

$\displaystyle \displaystyle \frac{x}{3-x}-\frac{2(3-x)}{3-x}<0$

$\displaystyle \displaystyle \frac{3x-6}{3-x}<0$

What I do is as follows:

1. Draw a number line. Make points on the number line at which the fraction will equal 0 or be undefined.

2. Substitute values into the equation from each region of the number line. If, in this case, a value turns out to be less than 0, then the entire region will work. Note that x=2, and x=3 do not work, because x=3 is undefined and x=2 will cause the fraction to evaluate to 0.

3. Once you have found regions that work, those are your limitations on x.
• Jan 31st 2011, 04:59 PM
Quote:

Originally Posted by jam2011
Book: Calculus with analytic geometry ( fourth edition )by protter and protter.

Problems(page 6 )

x/(3-x) < 2
= x< 2(3-x) by distribution

only if x<3

= x < 6-2x by adding 2x both side
= 3x < 6 by dividing 3 both side

please check my work if i done correctly. I'm little confuse when can i use cases in inequalities. Thank you in advance

$\displaystyle \displaystyle\ 2-\frac{x}{3-x}>0$

$\displaystyle \displaystyle\frac{6-2x}{3-x}-\frac{x}{3-x}>0\Rightarrow\frac{6-3x}{3-x}>0$

$\displaystyle \displaystyle\Rightarrow\ 3\left(\frac{2-x}{3-x}\right)>0$

This is positive if we have $\displaystyle \frac{(+)}{(+)}$ or $\displaystyle \frac{(-)}{(-)}$

which gives $\displaystyle -\infty<x<2$ and $\displaystyle 3<x<\infty$
• Jan 31st 2011, 05:08 PM
jam2011
Quote:

$\displaystyle \displaystyle\ 2-\frac{x}{3-x}>0$

$\displaystyle \displaystyle\frac{6-2x}{3-x}-\frac{x}{3-x}>0\Rightarrow\frac{6-3x}{3-x}>0$

$\displaystyle \displaystyle\Rightarrow\ 3\left(\frac{2-x}{3-x}\right)>0$

This is positive if we have $\displaystyle \frac{(+)}{(+)}$ or $\displaystyle \frac{(-)}{(-)}$

which gives $\displaystyle -\infty<x<2$ and $\displaystyle 3<x<\infty$

what i'm going to do to get this answer? thanks
• Jan 31st 2011, 05:17 PM
jam2011
it less than or greater than?
• Jan 31st 2011, 05:22 PM
jam2011
thank you for your information that i cannot multiply directly when the denominator has a variable...
• Jan 31st 2011, 05:32 PM
Quote:

Originally Posted by jam2011
what i'm going to do to get this answer? thanks

$\displaystyle \displaystyle\frac{2-x}{3-x}>0$

If the numerator is positive, then $\displaystyle x<2$

Notice the denominator is also positive for $\displaystyle x<2$

the denominator is negative if $\displaystyle x>3$

Notice that the numerator is also negative when $\displaystyle x>3$

In between these x, the numerator and denominator have opposite sign,
aside from when x=2 or 3.
• Jan 31st 2011, 05:39 PM
jam2011
can i use this:
since we have 3x-6/3-x<0 then

case 1: suppose x is positive, x>0
3x-6/3-x>0
3x-6>0 by multiplying 3-x both side
3x>6 by adding 6 both side
x>2

Case 2 : when x is negative then x<0

3x-6/3-x<0
3x-6>0 by multiplying 3-x both side
3x>6 by adding 6 both side
x>2

i use this pattern since this my instructor discussion? or i cannot apply it this time
• Jan 31st 2011, 05:42 PM
Prove It
First thing you should note is that $\displaystyle \displaystyle x \neq 3$ because that will give a $\displaystyle \displaystyle 0$ denominator.

$\displaystyle \displaystyle \frac{x}{3-x} < 2$

$\displaystyle \displaystyle \frac{x - 3 + 3}{3-x} < 2$

$\displaystyle \displaystyle \frac{3}{3-x} - \frac{3-x}{3-x} < 2$

$\displaystyle \displaystyle \frac{3}{3-x} - 1 < 2$

$\displaystyle \displaystyle \frac{3}{3-x} < 3$

$\displaystyle \displaystyle \frac{1}{3-x} < 1$.

In order for this inequality to hold, we require that the denominator be larger than the numerator, or for the denominator to be negative.

Case 1: $\displaystyle \displaystyle 3 - x > 1$

$\displaystyle \displaystyle 3 > x + 1$

$\displaystyle \displaystyle 2 > x$.

Case 2: $\displaystyle \displaystyle 3 - x < 0$

$\displaystyle \displaystyle 3 < x$.

So the solution is $\displaystyle \displaystyle x \in (-\infty, 2) \cup (3, \infty)$.
• Jan 31st 2011, 05:45 PM
jam2011
im confuse Mr. Archie Meade since ive posted that is less than sign but then your doing it in greater than...isn"t it greater ?or less than
• Jan 31st 2011, 05:46 PM
@ Prove It

Or for 3-x to be negative.
• Jan 31st 2011, 05:49 PM
Quote:

Originally Posted by jam2011
im confuse Mr. Archie Meade since ive posted that is less than sign but then your doing it in greater than...isn"t it greater ?or less than

You can do it either way.

For example x-2>0 is the same as 2-x<0

$\displaystyle \frac{x}{3-x}<2\Rightarrow\ 2>\frac{x}{3-x}$
• Jan 31st 2011, 05:53 PM
Prove It
Quote:

@ Prove It

Or for 3-x to be negative.

Yes, very true. Editing now...
• Jan 31st 2011, 06:32 PM
jam2011
Quote:

Originally Posted by Prove It
First thing you should note is that $\displaystyle \displaystyle x \neq 3$ because that will give a $\displaystyle \displaystyle 0$ denominator.

$\displaystyle \displaystyle \frac{x}{3-x} < 2$

$\displaystyle \displaystyle \frac{x - 3 + 3}{3-x} < 2$

$\displaystyle \displaystyle \frac{3}{3-x} - \frac{3-x}{3-x} < 2$

$\displaystyle \displaystyle \frac{3}{3-x} - 1 < 2$

$\displaystyle \displaystyle \frac{3}{3-x} < 3$

$\displaystyle \displaystyle \frac{1}{3-x} < 1$.

In order for this inequality to hold, we require that the denominator be larger than the numerator, or for the denominator to be negative.

Case 1: $\displaystyle \displaystyle 3 - x > 1$

$\displaystyle \displaystyle 3 > x + 1$

$\displaystyle \displaystyle 2 > x$.

Case 2: $\displaystyle \displaystyle 3 - x < 0$

$\displaystyle \displaystyle 3 < x$.

So the solution is $\displaystyle \displaystyle x \in (-\infty, 2) \cup (3, \infty)$.

This is in different way of solving this problem...i Like it.. Thank you very much
• Jan 31st 2011, 07:14 PM
jam2011
as I study your work in case 2, you have 3 - x < 0. Why is it that you used < 0 instead of <1(same as case 1) please help me explain me further.thanks a lot
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