rational equation and rearraging formaul

**dadon** · April 13th 2006, 02:15 AM

hey guys! my first post!

i needed some help

how do you so the following rational equation

(x-1)^2 / x-2 = 9 / 2

and also

how do you make y the subject of the formula when

y+2 / y+1 = x

i been trying to solve this all morning!

anyways guys if you can help please do..

.. oh yeah and a very good site for MATHS help

http://www.mathtutor.ac.uk/

http://www.mathcentre.ac.uk/

hope you find them useful

**topsquark** · April 13th 2006, 03:00 AM

Originally Posted by dadon

how do you so the following rational equation

(x-1)^2 / x-2 = 9 / 2

$\frac{(x-1)^2}{x-2}=\frac{9}{2}$

No factors cancel, so assuming x is not equal to 2, we can multiply both sides by (x-2) and multiply both sides by 2:

$2(x-1)^2=9(x-2)$
$2x^2-4x+2=9x-18$
$2x^2-13x+20=0$
$(2x-5)(x-4)=0$

So $2x-5=0$ or $x-4=0$

Finally: x = 5/2 or x = 4. Quick check: Neither of these says x = 2, so we're okay.

-Dan

**CaptainBlack** · April 13th 2006, 03:05 AM

Originally Posted by dadon

hey guys! my first post!

i needed some help

how do you so the following rational equation

(x-1)^2 / x-2 = 9 / 2

$ \frac{(x-1)^2} { x-2} = \frac{9}{ 2} $

so multiplying both sides by $x-2$ :

$ (x-1)^2 = \frac{9}{ 2} (x-2) $ ,

expanding the square:

$ x^2-2x+1 = \frac{9}{ 2} (x-2) $ ,

bringing everthing over to the LHS

$ x^2-\frac{13}{2}x+10 = 0 $ .

Which can now be solved using the quadratic formula.

RonL

**topsquark** · April 13th 2006, 03:05 AM

Originally Posted by dadon

how do you make y the subject of the formula when

y+2 / y+1 = x

$\frac{y+2}{y+1}=x$

Multiplying both sides by (y+1):

$y+2 = x(y+1)$
$y+2=xy+x$
$y-xy=x-2$
$y(1-x)=x-2$
$y = \frac{x-2}{1-x}$

This is sufficient, though some people might prefer the following form, given that x-1 = -(1-x):
$y =- \frac{x-2}{x-1}$

-Dan

**dadon** · April 13th 2006, 03:14 AM

thanks guys exactly what i needed!

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