What are the solutions to the inequality?

[Darth Frodo]

Homework Statement

\frac{3x + 1}{2x - 6} < 3

Homework Equations

The Attempt at a Solution

\frac{3x + 1}{2(x -3} < 3

\frac{3x +1}{x - 3} < 6

Assume x < 3

3x + 1 > 6(x - 3)
3x + 1 > 6x - 18
3x + 1 - 6x + 18 > 0
19 > 3x
x < 19/3

No Contradiction.


Assume x > 3
3x + 1 < 6x - 18
3x - 19 > 0
3x > 19
x > 19/3

No Contradiction




How do I know which set of values to take?

 

[uart]

Homework Statement

\frac{3x + 1}{2x - 6} < 3

Homework Equations

The Attempt at a Solution

\frac{3x + 1}{2(x -3} < 3

\frac{3x +1}{x - 3} < 6

Assume x < 3

3x + 1 > 6(x - 3)
3x + 1 > 6x - 18
3x + 1 - 6x + 18 > 0
19 > 3x
x < 19/3

No Contradiction. Assume x > 3
3x + 1 < 6x - 18
3x - 19 > 0
3x > 19
x > 19/3

No Contradiction

How do I know which set of values to take?
[

Hi Darth. You need to consider both the assumption and the consequent result in determining the solution region.

For example {x < 3 and x < 19/3} means that x must be less than 19/3, but also less than 3, but this is equivalent to simply stating x < 3.

 

[uart]

For your second solution region, x must be greater than 3, but also greater than 19/3. This is equivalent to simply stating x &gt; \ldots. (You fill that one in. )

 

[Darth Frodo]

I attempted a different method

Is the correct answer

x < 3
x > 19/3

 

[uart]

Yes that is correct.

BTW. What was your "different method". One alternative is to multiply both sides by the square of the denominator, which makes it into a single quadratic inequality.

 

[Darth Frodo]

Yeah, that was what I did
Then tested the point

 

[CAF123]

Could also do \frac{3x+1 - 3(2x-6)}{2x-6}&lt;0 simplify and then draw up an algebraic table to see when each of the quantities (3x-19), (2x-6) are negative, zero or positive (or undefined)for values of x less than 3, 3, 3<x<19/3, 19/3 and greater than 19/3.