In equalities

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January 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
x < 2 answer

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

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

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

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

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.
January 31st 2011, 04:59 PM
Archie Meade

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
x < 2 answer

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

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

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

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

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

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

Quote:

Originally Posted by Archie Meade

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

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

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

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

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

what i'm going to do to get this answer? thanks
January 31st 2011, 05:17 PM
jam2011

it less than or greater than?
January 31st 2011, 05:22 PM
jam2011

thank you for your information that i cannot multiply directly when the denominator has a variable...
January 31st 2011, 05:32 PM
Archie Meade

Quote:

Originally Posted by jam2011

what i'm going to do to get this answer? thanks

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

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

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

the denominator is negative if $x>3$

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

In between these x, the numerator and denominator have opposite sign,
aside from when x=2 or 3.
January 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
January 31st 2011, 05:42 PM
Prove It

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

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

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

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

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

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

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

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: $\displaystyle 3 - x > 1$

$\displaystyle 3 > x + 1$

$\displaystyle 2 > x$ .

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

$\displaystyle 3 < x$ .

So the solution is $\displaystyle x \in (-\infty, 2) \cup (3, \infty)$ .
January 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
January 31st 2011, 05:46 PM
Archie Meade

@ Prove It

Or for 3-x to be negative.
January 31st 2011, 05:49 PM
Archie Meade

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

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

Quote:

Originally Posted by Archie Meade

@ Prove It

Or for 3-x to be negative.

Yes, very true. Editing now...
January 31st 2011, 06:32 PM
jam2011

Quote:

Originally Posted by Prove It

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

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

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

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

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

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

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

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: $\displaystyle 3 - x > 1$

$\displaystyle 3 > x + 1$

$\displaystyle 2 > x$ .

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

$\displaystyle 3 < x$ .

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

This is in different way of solving this problem...i Like it.. Thank you very much
January 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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