# Homework Help: Isolating a variable when variable is in both sides of equation

1. Nov 12, 2012

### Sn0man

1. The problem statement, all variables and given/known data
y=(2-x)/(5x)

Solving for x=f(y)

I can get as far as 5xy=(2-x)

Tried 5xy=-1(x+2)

All attempts at solving this have given me the wrong answer, and I end up with x on both sides of the equation.

Stuck and not sure what to do.

Correct answer is x=2/(5y+1) , but I don't know how to get there.

Thanks for any help!

2. Nov 12, 2012

### symbolipoint

Start with what you know and go step by step.

From the given equation:
1. Multiply both members by 5x.
2. Add the additive inverse of you-know-what to both sides.
3. Can you then see to combine two terms of the variable, x ?
4. ...And if you did, then you can finish, finding x as a formula of y.

3. Nov 12, 2012

### HallsofIvy

This the same as 5xy= 2- x, you don't need the parentheses. If you want x only on the left, then add x to both sides.

4. Nov 12, 2012

### Bonaparte

The best way to solve this problem would be to separate the division part, that is to say 2/(5x)- x/(5x) = 2/(5x) - 1/5, then multiply all by 5x to get
5xy=2-x. Now add x to both sides to get 5xy+x = 2. Now do you notice something? We can factor this to x(5y+1)=2, divide by 5y+1 to get x=2/(5y+1)

Bonaparte

5. Nov 12, 2012

### Sn0man

Aha! Of course...thats what I was missing! I was so close just forgot to factor the x out.

Thanks very much - this problem was really getting under my skin!

6. Nov 12, 2012

### Mentallic

Separating the division is both an unnecessary step and makes things harder.

7. Nov 12, 2012

### SammyS

Staff Emeritus
An alternate method:
$\displaystyle y=\frac{2-x}{5x}$

$\displaystyle 5y=\frac{2-x}{x}$

$\displaystyle 5y=\frac{2}{x}-1$

$\displaystyle 5y+1=\frac{2}{x}$

$\displaystyle x=\frac{2}{5y+1}$​

8. Nov 12, 2012

### symbolipoint

NOT necessarily, in case it helps the o.p. to understand.

9. Nov 12, 2012

### SammyS

Staff Emeritus
Good point.

In this case, breaking up the rational expression (separating the division) leaves only one term with x in it -- a fairly uncomplicated term at that. This makes it rather easy to solve for x.

10. Nov 12, 2012

### Mentallic

I wouldn't think to explain to a student that's working with functions and factorizing that (a+b)/c is the same as a/c+b/c, but I suppose I shouldn't make that assumption.

11. Nov 12, 2012

### Bonaparte

Mentallic, not that you too seperated the division part, just later.
That was from (2-x)/x to 2/x-x/x. You are right that it is not necessary, but when he gets to much more complicated problems, I personally find it helpful to seperate. Where you seperated you could have multiplied by x both sides and continue, you are perfectly right that it is not "necessary".

Bonaparte

12. Nov 13, 2012

### symbolipoint

Actually, when working the problem forward again myself, Mentallic is right, that splitting the rational expression into a sum (or a difference) should not be necessary. The problem can be solved in four steps (slightly differently than #7, SammyS). Myself, I found no use in splitting the expression into sum or difference.

13. Nov 13, 2012

### Bonaparte

I'm tired of repeating myself, of course it is not necessary, I disagree it makes things harder however. I will rather spend a extra 30 seconds to separate then have the O.P. not understand and explain. Please use your time more efficiently and help others.

Thanks, Bonparte