# Help to solve this simple equation.

1. Apr 18, 2015

### ElectroViruz

Question : [ x / (x - 3) ] + 3 = 3 / (x - 3)

Attempt :

So what I did was simplifying the left hand side of the equation,

[ x / (x -3) ] + (3 / 1) = 3 / (x - 3)

[(4x - 9) / ( x - 3) ] = 3 / (x - 3)

then bring the (x - 3) from the right hand side to the left hand side to multiply

[ (4x - 9) / ( x - 3) ] . ( x - 3) = 3

Then the (x -3) cancels off

[ (4x - 9) / (x - 3) ] . (x - 3) = 3

Hence

4x - 9 = 3

Ans : x = 3

But that's not all

I discussed with my friends, some of them had the same answer, and some of them said there wasn't a solution to this. Those to who said there wasn't a solution, their working was

Question : [ x / (x - 3) ] + 3 = 3 / (x - 3)

Switch the position of 3 and right hand side,

[ x / (x - 3) ] - [ 3 / (x - 3) ] = - 3

Therefore,

[ (x - 3) / (x - 3) ] = -3

Hence there wouldn't be any solution.
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So which working is right ? x = 3 or no solution ?

Thanks in advance. I really appreciate it :)

2. Apr 18, 2015

### SteamKing

Staff Emeritus
If x = 3 really is a solution, what happens if you plug this value into the original equation?

3. Apr 18, 2015

### ElectroViruz

OH SHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH. OMFG. THANKS. I CAN'T BELIEVE I DIDN'T SEE THAT. KILL ME PLS.

EDIT : Oh umm umm, so that would mean that it's possible to bring a fraction to the other side like that ? *Haven't really done that in high school maths.

4. Apr 18, 2015

### Staff: Mentor

Here's the equation in a little nicer form:
$$\frac x {x - 3} + 3 = \frac 3 {x - 3}$$
When you're working with equations that involve rational expressions, the first thing you should do is note any potential solutions that cause division by zero. Clearly we can't end up with x = 3 here.

When you multiply both sides by (x - 3) you are tacitly assuming that x ≠ 3. If x were equal to 3, you would be multiplying both sides of the equation by zero, which is never a useful thing to do.

5. Apr 18, 2015

### Staff: Mentor

One other thing: Please don't delete the homework template.

6. Apr 18, 2015

### ElectroViruz

Ohhhhhhhhh I get it now. Thanks a lot, Mark :D

Sorry bout that >< I thought that was a guide line to post :x

7. Apr 18, 2015

### Stephen Tashi

You can test whether x=3 is a solution by substituting 3 for x in the original equation. It isn't a solution since the substitution creates expressions involving division by zero.

Technically, when you do algebraic manipulations as a logical demonstration, you are supposed to include some verbal statements in the work. For example, replacing $\frac{x}{x-3} + 3$ by $\frac{x}{x-3} + \frac{3(x-3)}{(x-3)}$ should be accompanied by the statement "if $x \ne 3$". So the final answer for the manipulations would be under the assumption that $x \ne 3$.

Most people don't heed such technicalities. They do the manipulations without stating the conditions for the manipulations to be correct and then check whether the answer really works after they have finished.

Your friends method should also include the comment "if $x \ne 3$ " for some of his manipulations. His work transforms the equation to another equation that has no solution, so the fact that he omitted the comment "if $x \ne 3$" didn't change his answer. However, neither of the methods is technically correct as a logical demonstration unless the comment "if $x \ne 3$" is included. Both methods give the same result if you include the comment. Both methods give the same result if you do manipulations without the comment and use substitution to check whether the answer they produce actually works as a solution to the original equation.

8. Apr 18, 2015

### ElectroViruz

Thank you. I was gonna ask how do I prove that there's no solution. So technically, switching the right hand side fraction to the left hand side isn't the solution, but making the statement beforehand by stating that the rational expression whereby x ≠ 3, then get my answer which is x = 3, which doesn't follow the statement, hence there isn't any solution. Am I right ?

9. Apr 18, 2015

### Ray Vickson

In this case, yes: there is no solution. If you write the equation as $f(x) = g(x)$, where
$$f(x) = \frac{x}{x-3} + 3, \; \text{and} \; g(x) = \frac{3}{x-3}$$
both $f(x)$ and $g(x)$ are undefined at $x = 3$, so $x = 3$ is not one of the allowed values of $x$. None of the allowed values (which are all $x \neq 3$) solve the equation.