1. ## inequalities problem

-3b<1/5

8a-8>7A-20

3-{3/2x-3}> 1/2(x+5)

2. Originally Posted by ms.butterfly

-3b<1/5

8a-8>7A-20

3-{3/2x-3}> 1/2(x+5)
$\displaystyle -3b<\frac{1}{5}$

Multiply by 5 and then divide by -15. Don't forget to reverse the inequality.

$\displaystyle -15b<1$

$\displaystyle b>-\frac{1}{15}$

Don't know what you want to do with the second one. It has two variables.

Is this the third one?

$\displaystyle 3-\left(\frac{3}{2}x-3\right)>\frac{1}{2}(x+5)$

3. yes, the second it I need addition and multiplication principles and yes that is the third one. I didn't know how to put it on the thread.

4. $\displaystyle 3-\left(\frac{3}{2}x-3\right)>\frac{1}{2}(x+5)$

First, remove parenthesis:

$\displaystyle 3-\frac{3}{2}x+3>\frac{1}{2}x+\frac{5}{2}$

Multiply every term by 2 to eliminate the fractions:

$\displaystyle 6-3x+6>x+5$

Combine terms:

$\displaystyle 12-3x>x+5$

Subtract x from both sides, and subtract 12 to both sides:

$\displaystyle -4x>-7$

Divide by -4 (Remember to reverse the inequality sign because you are dividing by a negative number)

$\displaystyle x<\frac{7}{4}$

Still not sure what you want in your second one.