# Thread: not sure if it is algebra...

1. ## not 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!?!

Thanks!!

2. Originally Posted by andyboy179
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!?!

Thanks!!
$\displaystyle \displaystyle \frac {2x+1} {3x-7}= 2$

multiply all with 3x-7

$\displaystyle 2x+1= 2\cdot (3x-7)$

$\displaystyle 2x+1 = 6x-14$

$\displaystyle 4x = 15 \Rightarrow x= 15/4$

3. if i multiply all by 3x-7 it would be, 6x+7/3x-7=2? then what would i do?

4. Originally Posted by andyboy179
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!?!

Thanks!!
Dear andyboy179,

First multiply both sides of the equation by 3x-7. Afterwards you can substract 2x from both sides. Hope you can continue.

5. so it would be, 6x+7/ 9x-49=2 then 4x+7/ 7x-49=2???

6. Originally Posted by andyboy179
if i multiply all by 3x-7 it would be, 6x+7/3x-7=2? then what would i do?
$\displaystyle \displaystyle \frac {2x+1} {3x-7}= 2 \Rightarrow \frac {(2x+1)\cdot (3x-7)} {(3x-7) }= 2\cdot (3x-7 ) \Rightarrow 2x+1= 2\cdot (3x-7)$

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 !!!

7. i really don't understand this!! could i do box method with 2x+1 and 3x-7? to get 6x^2-x-6?

8. Originally Posted by andyboy179
i really don't understand this!! could i do box method with 2x+1 and 3x-7? to get 6x^2-x-6?
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

$\displaystyle \frac {2x}{2}\cdot 2 = 1 \cdot 2 \Rightarrow \frac {2x \cdot 2 }{2} = 1\cdot 2 \Rightarrow 2x = 2 \Rightarrow x =2/2 = 1$

9. i kinda understand now. you say multiply both sides with 2 so how would i do 2x+1x2? and 3x-7x2?

10. Originally Posted by andyboy179
i kinda understand now. you say multiply both sides with 2 so how would i do 2x+1x2? and 3x-7x2?
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 ) you multiply both sides with THAT in your case it's 3x-7

so you will be free of the fraction

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

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

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

$\displaystyle \displaystyle \frac {x+y}{a+b} \cdot \frac {10}{3} = \frac {10x+10y}{3a+3b}$

11. OHHH so would it be 6x^2 - 10x -6?

12. Originally Posted by andyboy179
OHHH so would it be 6x^2 - 10x -6?
what would that be ... ????

lol... don't know what to do now .... probably someone who write better in English can help you maybe that's the problem

sorry...

13. 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!

14. Originally Posted by andyboy179
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!
you have solution in my first post (#2)

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

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