# Thread: am i doing this fraction correctly??

1. ## am i doing this fraction correctly??

hi this is my question:

when i times the bottom two i get 2x^2+x-3

then i times 3 by 2x+3 and 2 by x-1, and over all from that i get this:

i then times the bottom of the fraction by 2x-2 which will took like, 6x+9= 4x^2+2x-2

is this correct SO FAR?
if not where am i going wrong

2. Originally Posted by andyboy179
hi this is my question:

when i times the bottom two i get 2x^2+x-3

Say what? What does this mean? Did you multiply the denominators? What for? You better multiply the whole equation

by $(x-1)(2x+3)$ , thus getting:

$3(2x+3)=2(x-1)\Longrightarrow 6x+9=2x-2\Longrightarrow$ ...continue from here

Tonio

then i times 3 by 2x+3 and 2 by x-1, and over all from that i get this:

i then times the bottom of the fraction by 2x-2 which will took like, 6x+9= 4x^2+2x-2

is this correct SO FAR?
if not where am i going wrong
.

3. Originally Posted by andyboy179
then i times 3 by 2x+3 and 2 by x-1, ....
That's all you need to do; your other steps not necessary...
3(2x + 3) = 2(x - 1)
6x + 9 = 2x - 2
4x = -11
x = -11/4
That's it...nothing else!!

SUGGESTION: use "I multiply" instead of "I times" ; your teacher will be "happier"!

4. i don't know what to do after that.
would i work it out so it looks like this:

if so i don't know what to do from then on.

5. No, that just says "1= 1" and that doesn't help at all.

There is no point in posting here if you will not at least try what people suggest.

Your original problem is
$\frac{3}{x-1}= \frac{2}{2x+3}$
You then multiply both sides by the product of the denominators (x-1)(2x+3) to get
$\frac{3}{x-1}(x-1)(2x+3)= \frac{2}{2x+3}(x-1)(2x+3)$

No, as tonio told you, the "x- 1" terms cancel on the right and the "2x+3" terms cancel on the left leaving
$3(2x+3)= 2(x-1)$
which is the same as
$6x+ 3= 2x- 2$

Can you solve that equation?

6. oh i understand now!

you would get 4x+5=0 X=1.25

7. NO. 4x + 5 = 0; then 4x = -5; x = -4/5
Anyhow, Halls typoed: should be 4x + 11 = 0

8. Originally Posted by Wilmer
NO. 4x + 5 = 0; then 4x = -5; x = -4/5
Anyhow, Halls typoed: should be 4x + 11 = 0
Wilmer you have a typo.

$\displaystyle 4x+5=0\Rightarrow x=-\frac{5}{4}$

9. Originally Posted by andyboy179
hi this is my question:

when i times the bottom two i get 2x^2+x-3

then i times 3 by 2x+3 and 2 by x-1, and over all from that i get this:

i then times the bottom of the fraction by 2x-2 which will took like, 6x+9= 4x^2+2x-2

is this correct SO FAR?
if not where am i going wrong
You really should ask yourself "Why am I taking the steps that I'm taking ?"

If two fractions are equal, the pattern is:

$\frac{1}{2}=\frac{2}{4}=\frac{3}{6}=\frac{4}{8}=\f rac{5}{10}=\frac{6}{12}=......$

$\frac{3}{5}=\frac{6}{10}=\frac{9}{15}=\frac{12}{20 }=......$

Notice that the denominator of the first fraction "times" the numerator of the second fraction
is equal to the numerator of the first fraction "times" the denominator of the second one.

In other words... $\frac{2}{4}=\frac{3}{6}\Rightarrow\ 2*6=3*4$

$\frac{3}{5}=\frac{6}{10}\Rightarrow\ 3*10=5*6$

and so on.

Even when you have "x" in the fractions, it's still true

$\displaystyle\frac{3}{x-1}=\frac{2}{2x+3}\Rightarrow\ 3(2x+3)=2(x-1)$

as Wilmer explained.

If you want to play with common denominators as you were doing,
it's more work, but you need to choose one of a few ways to continue.

One way is to say "2 things are equal so we get zero by subtracting them".

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

Now you can do your "cross-multiplication stuff"

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

which means the numerator is zero

$3(2x+3)-2(x-1)=0\Rightarrow\ 6x+9-2x+2=0$

Of course, cross multiplication is just the following:

$\displaystyle\frac{3}{x-1}-\frac{2}{2x+3}=0\Rightarrow\frac{3}{x-1}\left[\frac{2x+3}{2x+3}\right]-\frac{2}{2x+3}\left[\frac{x-1}{x-1}\right]=0$

and again the numerator is zero.

Another way is to not bother subtracting at all:

$\displaystyle\frac{3}{x-1}=\frac{2}{2x+3}\Rightarrow\frac{3}{x-1}\left[\frac{2x+3}{2x+3}\right]=\frac{2}{2x+3}\left[\frac{x-1}{x-1}\right]$

and again, since the denominators are equal, the numerators must also be equal,
so we again obtain

$3(2x+3)=2(x-1)$

10. Originally Posted by dwsmith
Wilmer you have a typo.
$\displaystyle 4x+5=0\Rightarrow x=-\frac{5}{4}$
That was on purpose...

11. would i multiply 3 by 2x+3 and x-1 by 2
= 6x+9=2x-2
= 4x-11
11/4= -2.75

12. Originally Posted by andyboy179
would i multiply 3 by 2x+3 and x-1 by 2
= 6x+9=2x-2
= 4x-11
11/4= -2.75
Your working is correct here, but be careful with the notation! Look where the equal signs are below:

$6x+9=2x-2$

$4x = -11$

$x = -\frac{11}{4}$

$=-2.75$

I know it's picky, but in certain exams, method marks may not be granted if your equations are not set out correctly, regardless of whether your solution is correct, so it's best to be as clear as possible. Obviously, here it's obvious what you're trying to communicate, but it's best to avoid getting into any nasty habits.