# simplify check!

• Nov 1st 2008, 12:20 AM
jvignacio
simplify check!
simplify

$\frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

$
\frac {(x-3)(x-2)}{(x-2)^2}
$

correct? thanks
• Nov 1st 2008, 12:56 AM
CaptainBlack
Quote:

Originally Posted by jvignacio
simplify

$\frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

$
\frac {(x-3)(x-2)}{(x-2)^2}
$

correct? thanks

No. It would make it easier for us to check your work if you posted some working.

CB
• Nov 1st 2008, 01:01 AM
jvignacio
Quote:

Originally Posted by CaptainBlack
No. It would make it easier for us to check your work if you posted some working.

CB

sorry mate. ok here it is:

$\frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

$\frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}$

then cross multiply then mulitply denominators and start cancelling out.

$
\frac {(x-3)(x-2)}{(x-1)^2}
$
• Nov 1st 2008, 01:40 AM
CaptainBlack
Quote:

Originally Posted by jvignacio
sorry mate. ok here it is:

$\frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}$

$\frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}$

then cross multiply then mulitply denominators and start cancelling out.

$
\frac {(x-3)(x-2)}{(x-1)^2}
$

I don't understand what you mean, but standard method of of adding fractions gives:

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

and simplify

CB
• Nov 1st 2008, 01:48 AM
jvignacio
Quote:

Originally Posted by CaptainBlack
I don't understand what you mean, but standard method of of adding fractions gives:

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

and simplify

CB

do i expand brackets to simplify or just cancel out the terms like top (x-3) and bottom (x-3) etc?
• Nov 1st 2008, 01:56 AM
mr fantastic
Quote:

Originally Posted by jvignacio
do i expand brackets to simplify or just cancel out the terms like top (x-3) and bottom (x-3) etc?

How would you simplify $\frac{3 \times 5 - 2}{3}$ .... ? Cancel the 3 .... $\frac{\not{3} \times 5 - 2}{\not{3}} = \frac{5 - 2}{1} = 3$ .... ?
• Nov 1st 2008, 02:06 AM
jvignacio
Quote:

Originally Posted by mr fantastic
How would you simplify $\frac{3 \times 5 - 2}{3}$ .... ? Cancel the 3 .... $\frac{\not{3} \times 5 - 2}{\not{3}} = \frac{5 - 2}{1} = 3$ .... ?

$
\frac{(x-3)-(x-2)}{(x-1)}
$

this okay?
• Nov 1st 2008, 03:08 AM
mr fantastic
Quote:

Originally Posted by jvignacio
$
\frac{(x-3)-(x-2)}{(x-1)}
$

this okay?

NO!

Quote:

Originally Posted by mr fantastic
How would you simplify $\frac{3 \times 5 - 2}{3}$ .... ? Cancel the 3 .... $\frac{\not{3} \times 5 - 2}{\not{3}} = \frac{5 - 2}{1} = 3$ .... ?

Do you really think that $\frac{3 \times 5 - 2}{3}$ is equal to 3!! I was trying to make a point. The point is that you can't do this. What you cancel has to be common to all of the numerator and all of the denominator before it can be cancelled!
• Nov 1st 2008, 04:07 AM
CaptainBlack
Quote:

Originally Posted by jvignacio
do i expand brackets to simplify or just cancel out the terms like top (x-3) and bottom (x-3) etc?

You can only cancel something which is a factor of both the top and bottom of such an expressions and $x-3$ is not a factor of the top.

What you do is expand the brackets at the top, collect like powers and factorise. Then if any of the factors on top are the same as a factor on the bottom you can canncel such facors.

CB
• Nov 1st 2008, 08:40 PM
jvignacio
Quote:

Originally Posted by CaptainBlack
You can only cancel something which is a factor of both the top and bottom of such an expressions and $x-3$ is not a factor of the top.

What you do is expand the brackets at the top, collect like powers and factorise. Then if any of the factors on top are the same as a factor on the bottom you can canncel such facors.

CB

for the top, i expanded it out and got $\frac {-2x-5}{(x-3)(x-2)(x-1)}$
• Nov 1st 2008, 08:45 PM
mr fantastic
Quote:

Originally Posted by CaptainBlack
I don't understand what you mean, but standard method of of adding fractions gives:

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

and simplify

CB

Quote:

Originally Posted by jvignacio
for the top, i expanded it out and got $\frac {-2x-5}{(x-3)(x-2)(x-1)}$

The numerator should be -2x + 5.