Please Explain How To Solve Algebra Problem

The problem is 2/3x-9=17. The back of the book tells me it is 39 but I don't know how to get that answer. Can someone please explain?

Sounds like you need to get someone to work with you on the basics of solving problems.

First, we want to get x on one side.
2/3x -9 + 9 = 17 + 9 [We add 9 to both sides. This still makes them equal!]
2/3x = 26 [9' and -9 canceled out, and 17+9 = 26]
2/3x * 3 = 26 * 3 [ We multiply both sides by 3. This still makes them equal!]
2x = 78 [ We multiply out, 2/3 * 3 = 2]
2x / 2 = 78 / 2 [Divide both by 2, so that we will be left with just x on the left]
x = 39

You want to solve for x.

Add 9 to both sides of equation and you get 2/3x = 26
I am assuming that the variable x is part of the numerator not denominator.
I would re-write it as 2x/3 = 26 (I am assuming the question lists x in the numerator)
Multiply both sides by 3 and get 2x = 78 and solve for x to get x = 39.

So, the equation is (2/3)x -9=17
The first thing you do is add 9 to both sides.
That gives you (2/3)x=26
Next, you will multiply both sides by 3, because that will turn 2/3 into 2.
So now you have 2x=78
Now all you have to do is divide both sides by two, and there's your answer.
x=39

(2/3)x - 9 = 17
(2/3)x = 26 {added 9 to both sides}
x = 26(3/2) {divided both sides by 2/3, multiplying by reciprocal, 3/2}
x = 39 {multiplied 26 by 3/2}

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