# Inequality homework problem

## Homework Statement

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

## 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

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

How do I know which set of values to take?

## The Attempt at a Solution

Related Precalculus Mathematics Homework Help News on Phys.org
uart

## Homework Statement

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

## 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

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

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 $x > \ldots$. (You fill that one in. )

I attempted a different method

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.

Yeah, that was what I did
Then tested the point

CAF123
Gold Member

Could also do $$\frac{3x+1 - 3(2x-6)}{2x-6}<0$$ 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.