Question

can someone please walk me through how you find the solution to a rational equation? I don't understand how to do it..

Answer 1

do you have an example problem?

Answer 2

yea

Answer 3

there are a few ways to tackle this problem. One way is to "get rid" of the fractions.

Step 1: multiply both sides (and all terms) of the equation by (2x+1)
can you do that ?

Answer 4

uh can u write that out like what u want me to do?

Answer 5

for example, the first term would be
\[ \frac{x}{2x+1} \cdot (2x+1) = \frac{x}{\cancel{(2x+10}} \cdot \cancel{(2x+1)}= x\]

Answer 6

the idea is anything divided by itself is 1, so (2x+1) divided by (2x+1) turns into 1

but you have to multiply all terms on both sides by 2x+1 to keep things equal

can you do the other terms ?

Answer 7

uh idk

Answer 8

first, write down the original equation
\[ \frac{x}{2x+1} - \frac{1}{4} = \frac{2}{2x+1} \]
can you do that ?

Answer 9

yes idid

Answer 10

if you can, then write (2x+1) times each term. just put (2x+1) next to each term

Answer 11

k

Answer 12

it should look like this
\[ \frac{x}{2x+1}\cdot (2x+1) - \frac{1}{4} \cdot (2x+1)= \frac{2}{2x+1} \cdot (2x+1)\]

Answer 13

ok

Answer 14

next, remember that
\[ \frac{x}{2x+1} \cdot \frac{(2x+1)}{1} - \frac{1}{4} \cdot \frac{(2x+1)}{1} = \frac{2}{2x+1} \cdot \frac{(2x+1)}{1} \]

if you see the same thing in the top and bottom you can cancel them
can you do that ?

Answer 15

(2x+1) in the top and bottom can be cancelled.

Answer 16

im lost

Answer 17

let's concentrate on the first term:
\[ \frac{x}{2x+1} \cdot (2x+1) \]
you multiply top times top and bottom times bottom. for a "whole number" you make its bottom a 1

\[ \frac{x}{2x+1} \cdot \frac{(2x+1)}{1} \]
now multiply top times top and bottom times bottom
\[ \frac{x (2x+1)}{(2x+1)}\]

notice that you have (2x+1) divided by itself. that is 1
you get
\[ \frac{x (2x+1)}{(2x+1)} = 1\]

Answer 18

ok

Answer 19

**\[ \frac{x (2x+1)}{(2x+1)} = x \]
not 1!

Answer 20

\[ \frac{x}{2x+1}\cdot (2x+1) - \frac{1}{4} \cdot (2x+1)= \frac{2}{2x+1} \cdot (2x+1) \]
\[ x - \frac{2x+1}{4} = \frac{2}{2x+1} \cdot (2x+1) \]

can you simplify the last term ?

Answer 21

i gtg lunch :/

Answer 22

do you see you can cancel (2x+1) from the top and bottom ? that leaves 2 by itself

Answer 23

Here is a video you can watch later
http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/rational_funcs_tutorial/v/rational-equations

Answer 24

thank you!