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Mathematics 5 Online
OpenStudy (anonymous):

What is the sum 1/g+2 +3/g+1 a. 3/g+3 b. g+3/(g+1)(g+2) c. 4g+7/(g+1)(g+2) d. 2g+3/(g+1)(g+2)

OpenStudy (phi):

what common denominator can you use ?

OpenStudy (anonymous):

g+3?

OpenStudy (phi):

no, the common denominator is not g+3 I would multiply the two denominators to get a common denominator

OpenStudy (phi):

notice most of your answer choices do that

OpenStudy (anonymous):

ok so its (g+1)(g+2)

OpenStudy (phi):

yes. next, what do you multiply the first fraction by so you get that denominator ? remember you multiply top and bottom by the same thing

OpenStudy (anonymous):

(1)(g+1)?

OpenStudy (phi):

yes, and you do that in the bottom also so that you get \[ \frac{g+1}{(g+1)(g+2) }\]

OpenStudy (phi):

now what about the second fraction?

OpenStudy (anonymous):

(3)(g+2)=3g+2?

OpenStudy (phi):

close. you have 3 "packages" of (g+2) which means you have 3 of everything inside the package in other words, redo this part :3(g+2) becuase it's not 3g+2.

OpenStudy (anonymous):

ok

OpenStudy (phi):

do you get the idea? 3(g+2) means the same as (g+2)+(g+2) + (g+2) how many g's is that ? how many 2's ?

OpenStudy (anonymous):

3 gs and 3 2s

OpenStudy (phi):

and you can write that as 3g+3*2 or 3g+2+2+2 but most people would write it 3g+6

OpenStudy (anonymous):

ok

OpenStudy (phi):

now both fractions have the same denominator. you can add their "tops"

OpenStudy (anonymous):

so it would be 4g+7/(g+1)(g+2)

OpenStudy (phi):

yes

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