# Math Help - In equalities

1. Because in Case 2 you require the denominator to be negative.

2. 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
= 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
Yes,
Prove It's method is very nice.

You could also consider this method to avoid making a lot of calculations....

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

Start by examining the denominator and realise that if $x>3$
the denominator is always negative.
The numerator will always be positive as $x>3$, so we have $\frac{(+)}{(-)}=-$

All negative numbers are less than 2, hence if $3

Now, examine the denominator being positive. This happens when $x<3$

When the denominator is positive, we can multiply both sides by the denominator

$x<2(3-x)\Rightarrow\ x<6-2x\Rightarrow\ x+2x<6\Rightarrow\ 3x<6\Rightarrow\ x<2$

Therefore, the other solution is $-\infty

This method also avoids the mess of multiplying across by a negative denominator and reversing inequality signs.

3. Originally Posted by jam2011
can i use this:
since we have (3x-6)/(3-x)<0 then

case 1: suppose x is positive, x>0

No! you should be supposing that the entire denominator is positive.
Probably the examples your instructor used had "x only" in the denominator

(3x-6)/(3-x)>0 you've inadvertently switched the inequality, by imagining x>3

(3x-6)/(3-x)<0

3x-6>0 by multiplying 3-x both side

true if x>3, as the denominator is negative and we reverse the inequality

3x>6 by adding 6 both side
x>2

However, x must also be >3, so x>3 means x>2 also

Solution is x>3

Case 2 : when x is negative then x<0

You need x<3 so that the denominator is positive

(3x-6)/(3-x)<0
3x-6>0 by multiplying 3-x both side.... no

You don't switch the inequality this time because you are not multiplying by a negative number.

3x>6 by adding 6 both side
x>2

3x-6<0
3x<6
x<2

i use this pattern since this my instructor discussion? or i cannot apply it this time
You need to be aware that you choose the denominator being positive or negative, not x.

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