# Thread: Algebra Equation with Fractions... Help!

1. ## Algebra Equation with Fractions... Help!

Hi there...

I really need to know where I've gone wrong with this one, could some one walk me through it? I hate fractions at the best of times and am really finding this difficult...

Solve for x:

I've probably gone about it the wrong way, so any pointers would be great...!

2. ## Cancelling + A shorter method

Hey,

I haven't fully reviewed your method yet, but it looks like you've made a cancelling error between the last two stages. With:

$\displaystyle \frac{2x^2-x-6}{x(2x-3)}$, you can't cancel an x because not everything in the numerator has an x to be cancelled.

If it was:
$\displaystyle \frac{2x^2-x-6x}{x(2x-3)}$, you could cancel the x because this would equal:

$\displaystyle \frac{x(2x-1-6)}{x(2x-3)}$, but in the original question, this is not the case.

Anyway, here's how I would solve it:

$\displaystyle \frac{x+2}{x}-\frac{2}{2x-3}=0$

Times both sides by $\displaystyle x$:

$\displaystyle x+2-\frac{2x}{2x-3}=0$

Times both sides by $\displaystyle 2x-3$:

$\displaystyle (x+2)(2x-3)-2x=0$

Can you finish from there?

3. Quacky, you may have just blown my algebra world apart!

"you can't cancel an x because not everything in the numerator has an x to be cancelled."

I also understand the part about multiplying both sides by x, and the fact the that the right hand side remains as 0 because anything multiplied by 0 is 0. No problem!

However, when:
"Times both sides by "
why does the $\displaystyle 2x-3$ remain as $\displaystyle 2x-3$ wouldn't that change (because it contains an $\displaystyle x$)

On the (my) second from last step (before I've gone wrong), can I multiply both sides by $\displaystyle x(2x-3)$ which would leave me with $\displaystyle 2x^2-x-6$ - I can then factorise and solve?

Is my working out correct up to that stage...?

Really appreciate your help, i've learnt quite a bit already! Thanks sent!

p.s. are there any instructions on this site on how to use the math tags?

4. OK, i think I have worked it out... Could someone check my working? Please bear with me, i'm going to try to use the math tags...

$\displaystyle \frac{x+2}{x}-\frac{2}{2x-3}=0$

$\displaystyle \frac{(x+2)(2x-3)}{x(2x-3)}-\frac{2x}{x(2x-3)}=0$

$\displaystyle x(2x-3)\times\left(\frac{(x+2)(2x-3)}{x(2x-3)}-\frac{2x}{x(2x-3)}\right) =0$

$\displaystyle (x+2)(2x-3)-2x=0$

$\displaystyle 2x^2-3x+4x-6-2x=0$

$\displaystyle 2x^2-x-6=0$

$\displaystyle (-2x-3)(-x+2)=0$

Therefore:
$\displaystyle (-x+2)=(x-2)=0$

so
$\displaystyle x=2!!!!!!!!$

Anybody shed any light on this... sure it's right!

5. Originally Posted by MaverickUK82
OK, i think I have worked it out... Could someone check my working? Please bear with me, i'm going to try to use the math tags...

$\displaystyle \frac{x+2}{x}-\frac{2}{2x-3}=0$

$\displaystyle \frac{(x+2)(2x-3)}{x(2x-3)}-\frac{2x}{x(2x-3)}=0$

$\displaystyle x(2x-3)\times\left(\frac{(x+2)(2x-3)}{x(2x-3)}-\frac{2x}{x(2x-3)}\right) =0$

$\displaystyle (x+2)(2x-3)-2x=0$

$\displaystyle 2x^2-3x+4x-6-2x=0$

$\displaystyle 2x^2-x-6=0$

$\displaystyle (-2x-3)(-x+2)=0$

e^(i*pi): fine up to here

Therefore:
$\displaystyle (-x+2)=(x-2)=0$

so
$\displaystyle x=2!!!!!!!!$

Anybody shed any light on this... sure it's right!
Be careful here, $\displaystyle -x+2 =0$ is the same as $\displaystyle x-2=0$ but this a special case, valid only when the RHS is 0.

In every other case $\displaystyle -x+2 = 2-x$

x=2 is a possible solution (we will check later)

However from your factorisation we note that $\displaystyle -2x-3=0$ is also a solution.$\displaystyle x = -\frac{3}{2}$

To test if these solutions are valid sub them into the original equation.
$\displaystyle x=2$ is a valid solution

$\displaystyle x = -\frac{3}{2}$ is also a valid solution

6. Originally Posted by MaverickUK82
OK, i think I have worked it out... Could someone check my working? Please bear with me, i'm going to try to use the math tags...

$\displaystyle \frac{x+2}{x}-\frac{2}{2x-3}=0$

$\displaystyle \frac{(x+2)(2x-3)}{x(2x-3)}-\frac{2x}{x(2x-3)}=0$

$\displaystyle x(2x-3)\times\left(\frac{(x+2)(2x-3)}{x(2x-3)}-\frac{2x}{x(2x-3)}\right) =0$
I don't know what you did here or why. Just keep track of your thought process when working fractions [a major problem for young people these days who don't get enough practice in such things.] You do the same as when dealing with ordinary fractions involving just numbers. In fact, algebra in this instance is just an extension of arithmetic. Algebra may be thought of as generalised arithmetic ...the same rules hold, more generally.

You have found a common denominator. Good. Now you work on the numerator: [I don't have the means yet to write nice equations as you do ...will do in time... so I'll use ASCII]

(x + 2)/x - 2/(2x-3) = 0
[(x+2)(2x - 3) - 2x]/[x(2x-3)] = 0

Now, so long as x NOT 0 or (2x-3) NOT 0, the numerator is zero...

(x+2)(2x - 3) - 2x = 0
2x^2 - x - 6 = 0

This is a quadratic to solve for x. You could use the quadratic formula, but it will easily factor, so why not try that? You will find two values for x that satisfy these conditions.

With regard to your question about "cancelling", there is no such thing. What you really do is to divide by equal factors. The principle is the reduction of fractions, a very important concept in later studies of similarity and so on. When you have numbers with added terms in the numerator and denominator, you do not "cancel", but reduce the fraction when the same number [or algebraic quantity] divides into both numerator [all of it] and denominator [all of it].

7. e^(i*pi)
Generous Contributor

e^(i*pi)
Generous Contributor

Thanks very much for your help guys.

I didn't notice it was a quadratic formula, I can't believe I missed it!

I am aware that when it = 0 it is a special case.

Thanks again guys for your help - much appreciated!

Just one thing I'd like to explain briefly:

You asked why I don't times the numerator and denominator by $\displaystyle x$.

Look at the example of:

$\displaystyle \frac{1}{2}$

That is not even an equation, so you wouldn't need to do this, but if you times both the top and bottom by $\displaystyle x$, you are left with:

$\displaystyle \frac{1x}{2x}$ and the $\displaystyle x$'s will cancel to give:

$\displaystyle \frac{1}{2}$ which is what we originally started with.

The reason is that you are clearly multiplying by $\displaystyle \frac{x}{x}$ which is the same as timesing by one. You haven't actually multiplied by $\displaystyle x$ at all.

However, if you times just the numerator by $\displaystyle x$, you are left with:

$\displaystyle \frac{1x}{2}$

You have timsed by $\displaystyle \frac{x}{1}$ which is obviously the same as timesing by x, which is what we had wanted to do to start.

9. Thanks again Quacky - understood!