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Mathematics 19 Online
OpenStudy (anonymous):

how can you find the ? in "((3x^2+8x+4)/(2x+1))((2x^2+3x+1)/?)=3x+2"

OpenStudy (anonymous):

put your unknown denominator to be y

OpenStudy (anonymous):

then y= ((3x^2+8x+4)/(2x+1))((2x^2+3x+1)/3x+2

OpenStudy (anonymous):

and...

OpenStudy (anonymous):

why did 3x+2 became the denominator??

OpenStudy (anonymous):

y= (x+2)(x+1) after simplification

OpenStudy (anonymous):

in an equation suppose 5x=3y, if you take something from the right hand side to the left hand side thn it changes its sign. in the example above if you take y to left hand side it changes to 5x/y=3

OpenStudy (anonymous):

(3x^2+8x+4)= (3x+2)(x+2) (2x^2+3x+1)= (2x+1)(x+1)

OpenStudy (anonymous):

[(3x+2)(x+2)/(2x+1)][(2x+1)(x+1)/3x+2]=y

OpenStudy (anonymous):

\[(\frac{(3x^2+8x+4)}{(2x+1)})(\frac{(2x^2+3x+1)}{?})=3x+2\]\[\frac{(3x+2)(x+2)}{(2x+1)} \times \frac{(2x+1)(x+1)}{?}=3x+2\]\[\frac{(3x+2)(x+2)}{(2x+1)} \times (2x+1)(x+1)=(3x+2) (?)\]\[\frac{\frac{(3x+2)(x+2)(2x+1)(x+1)}{(2x+1)}}{(3x+2)}=(?)\]\[? = \frac{(3x+2)(x+2)(2x+1)(x+1)}{(2x+1)(3x+2)}\] Now, cross out the common factors and get your answer!

OpenStudy (anonymous):

did you understand?

OpenStudy (anonymous):

sort of

OpenStudy (anonymous):

k.. if you want further help i cud e-mail you some scanned texts to help you better understand

OpenStudy (anonymous):

really??? my e-mail is viancarodolfo@yahoo.com

OpenStudy (anonymous):

i will send you the texts after a around 3 hours or so.. is it k???

OpenStudy (anonymous):

its wonderful,, im already failing math so this really helps me a lot

OpenStudy (anonymous):

don't worry i will help u out as much i can

OpenStudy (anonymous):

actuly i m doing my honors in mathematics.. so i wud b happy to help u

OpenStudy (anonymous):

thanks so much in advance

OpenStudy (anonymous):

you are welcome

OpenStudy (anonymous):

SORRY NET WAS NOT WORKING

OpenStudy (anonymous):

as soon as i finsh my writing, m scanning it to you

OpenStudy (anonymous):

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