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Mathematics 21 Online
OpenStudy (anonymous):

Just help with 2 problems please! 1. What is the simplified form of (5x+4)/(x+7) - (2x+1/(x+7)? a.(3x+3)/(x+7) b.(3x+5)/(x+7) c.(7x+3)/(x+7) d.(7x+5)/(x+5) 2. What is the simplified form of (y^2+7y+12)/(y^2 + 8y +15)? a.y-4/y-5 b.y-4/y+5 c.y+4/y-5 d.y+4/y+5

OpenStudy (anonymous):

Some one pleaseeeeee help me !

OpenStudy (amistre64):

well, the first follows the same concept as:\[\frac ab - \frac cb = \frac{a-c}{b}\]

OpenStudy (amistre64):

the second one is a little more complicated, but youd have to know how to factor quadratics

OpenStudy (anonymous):

so would the first one be D?

OpenStudy (anonymous):

actually I think It would be A correct??

OpenStudy (amistre64):

A does look better to me

OpenStudy (anonymous):

thank you ! how about the second one ??

OpenStudy (amistre64):

do you know how to multiply binomials? alot of places use a method they call "FOIL"

OpenStudy (anonymous):

no I don't know how to do that

OpenStudy (amistre64):

you must have something you can fall back on; they dont give you these problems in a vaccuum

OpenStudy (amistre64):

if we try to conform this to an option we would have be faced with trying to fit a common factor like this \[\frac{y^2+7y+12}{y^2 + 8y +15}=\frac{(y-4)(y+?)}{(y-5)(y+?)}\]

OpenStudy (amistre64):

using each option of course in turn

OpenStudy (amistre64):

since the last term is equal to the product of the ... last things .... 12 = -4*? 15 = -5*? in both cases -3 is a solution; but thats not a guarentee

OpenStudy (amistre64):

a and d have the same chances of being correct by that method

OpenStudy (amistre64):

so we look to the middle term; +7y has to be the addition of the last two thingys ... is -4-3 = 7? no so that leaves me to believe that it is not A

OpenStudy (anonymous):

Thank you so much I think I understand how to do it now!

OpenStudy (amistre64):

there are more mathical ways to solve it for sure; but they require you to have studied the material that these question relate to first :) good luck

OpenStudy (anonymous):

Thank you :)

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