Simplifying Algebraic Fractions...

Question:

$\frac{2x-3}{6} + \frac{x+2}{3} = \frac{5}{2}$

After cross multiplying the first two fractions, I got:

$\frac{6x-9+6x+12}{18} = \frac{5}{2}$

yes that is correct and your working is fine

Right, you came about this the wrong way, initially you must cross multiply but do not forget to multiply ALL the factors in the question.

So if you start with:

$\frac{2x-3}{6} + \frac{x+2}{3} = \frac{5}{2}$

you need to multiply everything by 6 first, so:

$2x+3 + \frac{6x+12}{3} = \frac{30}{2}$

Simplifying to give me:

Then divide $6x+12$ by 3 which gives you:
$2x+4$

which gives you: $2x+3+2x+4=15$

Which is $4x+7 = 15$
So $4x = 8$
And $x=2$

Right, you came about this the wrong way, initially you must cross multiply but do not forget to multiply ALL the factors in the question.

So if you start with:

$\frac{2x-3}{6} + \frac{x+2}{3} = \frac{5}{2}$

you need to multiply everything by 6 first, so:

$2x+3 + \frac{6x+12}{3} = \frac{30}{2}$

Simplifying to give me:

Then divide $6x+12$ by 3 which gives you:
$2x+4$

which gives you: $2x+3+2x+4=15$

Which is $4x+7 = 15$
So $4x = 8$
And $x=2$

Who says it has be factorised. Complete the square or use the equation

Sorry, I have one more question which im not sure if my working is correct.

Question:

$\frac{1}{x} + \frac{3}{2-x} = 0$

Now would my working out be correct for this question:

$\frac{1(2-x) + 3(x)}{x(2-x)}$
Which gives me:
$\frac{-x+2+3x}{-x^2+2x}$

Which when I move the bottom to the other side, gives me:

$2x+2 = x^2 - 2x$

And so when I rearrange this, I get:

$x^2 -4x -2$

And I cant factorise this

So Can someone point out where I've gone wrong please. I have a feeling I messed up at the beggining :/

Thanks

How did you move it? By multiplying if so thats wrong