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Mathematics 16 Online
OpenStudy (anonymous):

Solve (x/3) + (x/7) < 2

OpenStudy (anonymous):

solve for x

OpenStudy (noseboy908):

remember the rules of adding fractions? they always have to have the same denominator. so what can you multiply the 3, then the 7 by to have the same denom?

OpenStudy (anonymous):

the 3 by 7 and the 7 by 3

OpenStudy (anonymous):

7x+3x<42 then 10x<42 x<4.2 thats the answer

OpenStudy (noseboy908):

ok, so what are you left with on top (on the left side)?

OpenStudy (anonymous):

7x

OpenStudy (noseboy908):

ok, and what about on top of the second fraction?

OpenStudy (anonymous):

3x

OpenStudy (noseboy908):

add them up

OpenStudy (anonymous):

so then it'd be 10x/21

OpenStudy (anonymous):

I understand it from here, I just didn't know how to start the problem :p

OpenStudy (noseboy908):

ok, hope I helped

OpenStudy (anonymous):

you did, thanks!!

OpenStudy (anonymous):

no just i said what to do

OpenStudy (haseeb96):

x/3 + x/7 < 2 Solution:- First we add these fractions whose sum is 10x/21 if we want to check the answer whether i am right or wrong you just divde 10\21 and its answer would 0.4761 0.4761 < 2 Hence proved

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