Solving an absolute value inequality

[physphys]

Homework Statement

lx/(x-2)l < 5

Homework Equations

The Attempt at a Solution

x/(x-2) < 5
x< 5x-10
10 < 4x
5/2 < x

x/(x-2) > -5
x > -5x+10
6x > 10
x > 5/3

The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?

 

[SammyS]

Homework Statement

lx/(x-2)l < 5

Homework Equations

The Attempt at a Solution

x/(x-2) < 5
x< 5x-10
10 < 4x
5/2 < x

x/(x-2) > -5
x > -5x+10
6x > 10
x > 5/3

The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?

For P a positive real number, $\ \displaystyle \left| f(x) \right|<P\$ is equivalent to $-P<f(x)<P\ .$

More to the point, $\displaystyle \ \frac{x}{x-2}<5\$ is not equivalent to $\ x<5(x-2)\ .\$ In fact, if x-2 is negative, then you need to change the sense of the inequality .


You're much better off to write $\displaystyle \ \frac{x}{x-2}<5\$ as $\displaystyle \ \frac{x}{x-2}-5<0\,,\$ then write the left-hand side as a single rational expression by using a common denominator .

 

[physphys]

For P a positive real number, $\ \displaystyle \left| f(x) \right|<P\$ is equivalent to $-P<f(x)<P\ .$

More to the point, $\displaystyle \ \frac{x}{x-2}<5\$ is not equivalent to $\ x<5(x-2)\ .\$ In fact, if x-2 is negative, then you need to change the sense of the inequality .


You're much better off to write $\displaystyle \ \frac{x}{x-2}<5\$ as $\displaystyle \ \frac{x}{x-2}-5<0\,,\$ then write the left-hand side as a single rational expression by using a common denominator .

Okay so I simplified that farther to achieve this, 2(3x-5) / x-2 > 0. So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?

 

[Mark44]

The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?

The answer can't possibly be "x < 5/3 and x > 5/2". There aren't any numbers that are simultaneously smaller than 5/3, and larger than 5/2.

Okay so I simplified that farther to achieve this, 2(3x-5) / x-2 > 0.

How did you get this? I'm assuming you are starting from x/(x - 2) - 5 < 0, and took Sammy's advice to combine the two terms on the left into a single rational expression.

Also, if 2(3x-5) / x-2 is supposed to mean this:
$$ \frac{2(3x - 5)}{x - 2}$$
then you need to write it with parentheses around the entire denominator, like so:
2(3x-5) / (x-2)
Otherwise, we would interpret this to mean this:
$$ \frac{2(3x - 5)}{x} - 2$$

So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?

 

[SammyS]

Okay so I simplified that farther to achieve this, 2([STRIKE] 3 [/STRIKE] 2x-5) / ( x-2 ) > 0. So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?

So you have one expression, 2(2x-5), divided by another, (x-2), and the result is positive. How does the sign of the numerator compare to the sign of the denominator ?