how can you find the ? in "((3x^2+8x+4)/(2x+1))((2x^2+3x+1)/?)=3x+2"
put your unknown denominator to be y
then y= ((3x^2+8x+4)/(2x+1))((2x^2+3x+1)/3x+2
and...
why did 3x+2 became the denominator??
y= (x+2)(x+1) after simplification
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
(3x^2+8x+4)= (3x+2)(x+2) (2x^2+3x+1)= (2x+1)(x+1)
[(3x+2)(x+2)/(2x+1)][(2x+1)(x+1)/3x+2]=y
\[(\frac{(3x^2+8x+4)}{(2x+1)})(\frac{(2x^2+3x+1)}{?})=3x+2\]\[\frac{(3x+2)(x+2)}{(2x+1)} \times \frac{(2x+1)(x+1)}{?}=3x+2\]\[\frac{(3x+2)(x+2)}{(2x+1)} \times (2x+1)(x+1)=(3x+2) (?)\]\[\frac{\frac{(3x+2)(x+2)(2x+1)(x+1)}{(2x+1)}}{(3x+2)}=(?)\]\[? = \frac{(3x+2)(x+2)(2x+1)(x+1)}{(2x+1)(3x+2)}\] Now, cross out the common factors and get your answer!
did you understand?
sort of
k.. if you want further help i cud e-mail you some scanned texts to help you better understand
really??? my e-mail is viancarodolfo@yahoo.com
i will send you the texts after a around 3 hours or so.. is it k???
its wonderful,, im already failing math so this really helps me a lot
don't worry i will help u out as much i can
actuly i m doing my honors in mathematics.. so i wud b happy to help u
thanks so much in advance
you are welcome
SORRY NET WAS NOT WORKING
as soon as i finsh my writing, m scanning it to you
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