but im not sure how i would solve this (question in attachment!)

could someone please point me in the right direction to solving this question!?!

Attachment 18912

Thanks!!

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- Sep 13th 2010, 06:22 AMandyboy179not sure if it is algebra...
but im not sure how i would solve this (question in attachment!)

could someone please point me in the right direction to solving this question!?!

Attachment 18912

Thanks!! - Sep 13th 2010, 06:36 AMyeKciM
- Sep 13th 2010, 06:40 AMandyboy179
if i multiply all by 3x-7 it would be, 6x+7/3x-7=2? then what would i do?

- Sep 13th 2010, 06:40 AMSudharaka
- Sep 13th 2010, 06:42 AMandyboy179
so it would be, 6x+7/ 9x-49=2 then 4x+7/ 7x-49=2???

- Sep 13th 2010, 06:47 AMyeKciM
$\displaystyle \displaystyle \frac {2x+1} {3x-7}= 2 \Rightarrow

\frac {(2x+1)\cdot (3x-7)} {(3x-7) }= 2\cdot (3x-7 ) \Rightarrow 2x+1= 2\cdot (3x-7) $

so you will lose fraction after dividing $\displaystyle \displaystyle \frac {(2x+1)\cdot (3x-7)}{(3x-7)} = (2x+1) $

do you understand ?

P.S. when said multiply all .... means that you multiply and left and right side of equation !!! - Sep 13th 2010, 06:50 AMandyboy179
i really don't understand this!! could i do box method with 2x+1 and 3x-7? to get 6x^2-x-6?

- Sep 13th 2010, 07:00 AMyeKciM
let's look at this this way ...

$\displaystyle 2x = 2 \Rightarrow x = 2/2 = 1 $

multiply both sides of equation with 2 and you will have

$\displaystyle 4x= 4 \Rightarrow x = 4/4 = 1 $

do you see that nothing is changed ?

now with fractions ....

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

because of that under fraction is 2 (can be anything) to lose that fraction you multiply both sides with that whatever is under fraction ....

in this case multiply with 2

$\displaystyle \frac {2x}{2}\cdot 2 = 1 \cdot 2 \Rightarrow \frac {2x \cdot 2 }{2} = 1\cdot 2 \Rightarrow 2x = 2 \Rightarrow x =2/2 = 1 $ - Sep 13th 2010, 07:07 AMandyboy179
i kinda understand now. you say multiply both sides with 2 so how would i do 2x+1x2? and 3x-7x2?

- Sep 13th 2010, 07:12 AMyeKciM
you don't multiply by two in your problem there... i just do that with my example to show you that nothing changes if you multiply both sides (or add or divide or ... ) Whatever is "down" in fraction (can't figure out how to write it on English :D) you multiply both sides with THAT :D in your case it's 3x-7 :D

so you will be free of the fraction :D

P.S. when multiplying fraction with number you don't multiply down expression :D

$\displaystyle \displaystyle \frac {x+y}{a+b} \cdot 10 = \frac {10x +10y}{a+b} $

but only if you are multiplying fraction with fraction ...

$\displaystyle \displaystyle \frac {x+y}{a+b} \cdot \frac {10}{3} = \frac {10x+10y}{3a+3b} $ - Sep 13th 2010, 07:14 AMandyboy179
OHHH so would it be 6x^2 - 10x -6?

- Sep 13th 2010, 07:20 AMyeKciM
- Sep 13th 2010, 07:22 AMandyboy179
would the answer be 2.3?? don't worry i can understand the english but im just finding it hard to pick up how to do these sort of questions!

- Sep 13th 2010, 07:32 AMyeKciM
you have solution in my first post :D (#2)

it's not problem for you with English ... it's mine :D:D:D

P.S. you need to get familiar with basic operations with fractions, and on fractions (as multiplying , divide, add , subtract .... ) so you will not have problem piking up stuff like this :D