linear inequality

[creationbaase]

Hello All! If someone could give me some help with this. I'm taking College Algebra this semester and was given some hw problems but I haven't taken math in a while. any help is appreciated

the topic section is called linear inequality

2/3 x (fraction) - 2 < 5/3 x (fraction) ( in each fraction x is next to the fraction not a #)

thanks all!

 

[Loren]

>2/3 x (fraction) - 2 < 5/3 x (fraction) ( in each fraction x is next to the fraction not a #)

I think you are saying...
\(\displaystyle \frac{1}{3}x-2<\frac{5}{3}x\)

If so, operate on the inequality using the same rules and procedures as if it were an equation, except when multiplying or dividing by a negative number, you must change the sense of the inequality.

First step --- Multiply both sides by 3.

Your move.

 

[creationbaase]

its 2/3 x

so if i multiply both sides by 3 than I would end up with

6/9x - 2 = 15/9x

am i going in the right direction?

 

[Loren]

\(\displaystyle 3\times\frac{2}{3}x = 2x\) and...

 

[mmm4444bot]

… am i going in the right direction?

No.

When we multiply two fractions, the numerator in the product comes from multiplying numerator times numerator. The denominator in the product comes from multiplying denominator times denominator.

(I hope that you know what these words mean.)

Shown symbolically, this process is:

A/B * C/D = (A * C)/(B * D)

Loren's instruction to multiply each term by 3 means multiply by the fraction 3/1.

3/1 * 2/3 = (3 * 2)/(1 * 3) = 6/3 = 2/1 = 2

Doing this eliminates all of the fractions in your original inequality. That's the goal of this first step.

3/1 * 5/3 = ?