solving rational expressions

[kathyjo82]

i'm in an intermediate algebra college math class.

i'm having trouble solving one problem for an assignment. i emailed my teacher, but not sure if or when i'll get a reply back.

the answer is five thirds. directions say to solve.

here is the equation.

X-2 X-1
---- = -----
X-3 X+1

(i'm sorry that didn't come out looking right, but i hope you get the idea.)

any help is greatly appreciated. thanks in advance.

 

[Unco]

G'day, kathyjo82!

We have

X-2      X-1
----  = -----
X-3      X+1

Clear out the denominator by multiplying both sides by (x-3)(x+1).

We now have:

(X-2)(X+1) = (X-1)(X-3)

These are quadratics, aren't they? So expand using FOIL or whatever you do:

X² - 2X + X - 2 = X² - 3X - X + 3

Simplify:

X² - X - 2 = X² - 4X + 3

Further simplify:

-X + 4X = 3 + 2

3X = 5

X = 5/3

 

[kathyjo82]

=]

thank you.

 

[Denis]

You may find this easier, Kathy:
if a/b = c/d
then ad = bc
Known as "cross multiplication".

(x/2) / (x-3) = (x-1) / (x+1)
Works same way:
(x-2)*(x+1) = (x-3)*(x-1)

After some practice, you'll find (well, I do anyway!) that it's a good idea
to rearrange an equation (if possible, of course) such that a cross multiplication
is possible: MUCH EASIER than a huge multiplication by the LCD.