I can see how a problem like this would give an introductory algebra student some trouble, especially if you don't know how to distribute fractions.
Lets focus in on the right hand side first. We have:
[tex] \frac{2}{3}(x - 9)[/tex]
What we want to do here is multiply x and -9 by 2/3. The reason why we doing this is because you can multiply a number next to a pair of parentheses into each term inside the parentheses. In this case, the terms inside the parentheses are x and -9, and the number we are multiplying by is actually a fraction (2/3). So we can write it like this:
[tex] \frac{2}{3} \times x + \frac{2}{3} \times - 9[/tex]
(2/3) * x is easy, all you have to do keep the fraction next to the x:
[tex] \frac{2}{3}x[/tex]
(2/3) * 9 is a different story though. If you recall how to multiply fractions from elementary, you should know that two fractions multiplied together is equal to the numerators multiplied over the denominators mulltiplied together. The problem here is that -9 is not a fraction. "So what do we do thehotdogman93?!"
You give up. There's no more hope.
No I'm kidding. You can actually write a 1 underneath the -9 to turn it into a fraction. It's going to look like this:
[tex] \frac{2}{3} \times \frac{ - 9}{1} [/tex]
Now you can multiply across. The 2 and -9 multiply together, and the 3 and 1 multiply together. This is what it is going to look like:
Congradulations, we have the right hand side now. Right now, the problem looks like this:
[tex]5 - 4x > \frac{2}{3}x - 6[/tex]
Now, there are a couple of different moves you can make here to solve for x, and all of them will give you the same answer, but I'm going to procede by adding 4x to both sides. By doing this, the -4x on the left will go away. On the right, we need to add (2/3)x and 4x together. More fraction math! This is what it looks like so far:
[tex]5 > \frac {2}{3}x + 4x - 6[/tex]
To add the the 4 and the (2/3) together, we need to use the same trick we used before, drawing a 1 underneath the 4 to turn it into a fraction. So we have:
[tex]\frac{2}{3} + \frac {4}{1}[/tex]
Now, we cant just add across in the same way that we multiplied across before. We actually need to find the least common denominator, in this case, it is 3. It should look like this:
Almost done! I promise. Our problem now looks like this:
[tex]5 > \frac{14}{3}x - 6 [/tex]
I'm going to add 6 to both sides so the -6 on the right goes away and we have something like this:
[tex]11 > \frac{14}{3}x[/tex]
The last thing here, is to get rid of the (14/3) next to the x, so the trick here is to multiply both sides by (3/14), which is the inverse of the fraction. x is going to be by itself on the right, on the left, we multiply 11 by (3/14) using the same process that we used to multipy -9 with (2/3).