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Mathematics 20 Online
OpenStudy (anonymous):

Solve

OpenStudy (anonymous):

OpenStudy (anonymous):

ick

OpenStudy (anonymous):

what

OpenStudy (anonymous):

it is ugly you have to add on the right first

OpenStudy (anonymous):

\[\frac{4x+1}{12}+\frac{2x+2}{2x+1}\] \[\frac{(4x+1)(2x+1)+12(2x+2)}{12(2x+1)}\] \[\frac{8x^2+30x+25}{12(2x+1)}\] then \[\frac{2x-1}{6}=\frac{8x^2+30x+25}{12(2x+1)}\]

OpenStudy (anonymous):

at least we can multiply both sides by 6 and get \[2x-1=\frac{8x^2+30x+25}{4x+2}\] then \[(2x-1)(4x+2)=8x^2+30x+25\] and now we are ready to solve

OpenStudy (anonymous):

\[8x^2-2=8x^2+30x+25\] \[-2=30x+25\] \[-28=30x\] \[x=-\frac{28}{30}=-\frac{14}{15}\] hhmm i wonder if there is an easier way, or if i made mistake

OpenStudy (anonymous):

damn i made a mistake!

OpenStudy (anonymous):

idk

OpenStudy (anonymous):

\[-2=30x+25\] \[-27=30x\] \[x=-\frac{27}{30}=-\frac{9}{10}\]

OpenStudy (anonymous):

Ha thats the answer I put .. I just guessed lol

OpenStudy (anonymous):

multiple choice?

OpenStudy (anonymous):

yupp

sam (.sam.):

you put 'solve' in-front of the expression , you can even get the steps :D

OpenStudy (anonymous):

very rectangular

sam (.sam.):

?

OpenStudy (anonymous):

@.Sam. true, but that is the way a machine solves it. i am not a human would have solved it that way

OpenStudy (anonymous):

notice all the fractions

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