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Algebra 22 Online
OpenStudy (anonymous):

divide and simplify to lowest terms 3y+5/6y+1 + 3y+5/y+6

OpenStudy (anonymous):

Starting with the form \[\frac{3y+5}{6y+1}+\frac{3y+5}{y+6}\] We multiply each expression by the conjugate of the other's denominator. For instance: For \[\frac{3y+5}{6y+1}\]We multiply by \[\frac{y+6}{y+6}\] We can do this because that expression equals 1. By doing this to both sides, we end up with a common denominator used to add the fractions. In full: \[\frac{3y+5}{6y+1}(\frac{y+6}{y+6})+\frac{3y+5}{y+6}(\frac{6y+1}{6y+1})\] It looks a bit ugly, but through distribution we get \[\frac{3y^2+23y+30}{6y^2+37y+6}+\frac{18y^2+33y+5}{6y^2+37y+6}\] Still ugly, but with equivalent denominators, we can add to get \[\frac{21y^2+56y+35}{6y^2+37y+6}\] Go ahead and factor out a 7 on top \[7\frac{3y^2+8y+5}{6y^2+37y+6}\] And then factor back out, since we can't cancel anything. \[7\frac{(y+1)(3y+5)}{(y+6)(6y+1)}\] I'm not a big fan of these problems either, don't worry

OpenStudy (anonymous):

idk why it's telling me its wrong. thats strange ugh i really do hate these problems

OpenStudy (anonymous):

,...?

OpenStudy (anonymous):

it says simplify AND type in factored form

OpenStudy (anonymous):

I'm thinking about it...if you wanted to separate them all out, you could get to \[\frac{9}{2 (6 y+1)}-\frac{13}{y+6}+\frac{7}{2}\]

OpenStudy (anonymous):

it said correct answer was y+6/6y+1

OpenStudy (anonymous):

unless the form you typed originally is wrong, that's not possible. One of the two would need to be reciprocated

OpenStudy (anonymous):

i copyed and pasted over idk want to try a different one? please

OpenStudy (anonymous):

yeah sure

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