# Algebra Problem

1. Oct 4, 2007

### kuahji

A problem in my textbook shows
25/9(20-y)^2 = 144+(20-y)^2
Then the very next step shows
16/9(20-y)^2=144

Maybe its just me, but I can't seem to figure out the algebra behind this step in the problem, any ideas on where to start? I thought about subtracting/dividing the (20-y)^2 from the right side, but that seems to be getting me nowhere.

2. Oct 4, 2007

### Dick

(20-y)^2=(9/9)(20-y)^2. You just need a common denominator.

3. Oct 4, 2007

### kuahji

Thanks, not sure why but I didn't even thing about the 9 being in the denominator in the first term. Probably because of the way it was written.

4. Oct 5, 2007

### arildno

No, the main reason why students overlook this sort of thing, is that they don't consider an expression like (20-y)^2 as JUST ANOTHER NUMBER, but as something far more etheral and mysterious.

5. Oct 5, 2007

### HallsofIvy

Staff Emeritus
If it was actually written 25/9(20-y)^2 in your book then I can see your confusion. What you give means
$$\frac{25}{9}(20-y)^2$$
but you might confuse it with
$$\frac{25}{9(20-y)^2}$$
I can see no good reason for not using the "clear" form in a printed text book.