# rational equation and rearraging formaul

• Apr 13th 2006, 02:15 AM
rational equation and rearraging formaul
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
• Apr 13th 2006, 03:00 AM
topsquark
Quote:

how do you so the following rational equation

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

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

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:

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

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

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

-Dan
• Apr 13th 2006, 03:05 AM
CaptainBlack
Quote:

hey guys! my first post!

i needed some help

how do you so the following rational equation

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

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

so multiplying both sides by $\displaystyle x-2$:

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

expanding the square:

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

bringing everthing over to the LHS

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

Which can now be solved using the quadratic formula.

RonL
• Apr 13th 2006, 03:05 AM
topsquark
Quote:

how do you make y the subject of the formula when

y+2 / y+1 = x

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

Multiplying both sides by (y+1):

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

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

-Dan
• Apr 13th 2006, 03:14 AM