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Mathematics 13 Online
OpenStudy (anonymous):

Solve these proportions. x/3 = x + 2/2 x=

sam (.sam.):

(x)/(3)=x+(2)/(2) x=3x+3 -2x=3 x=-(3)/(2)

sam (.sam.):

\[\huge OR\] (x)/(3)=(x+2)/(2) x=(3(x+2))/(2) 2x=(3(x+2))/(2)*2 2x=3(x+2) 2x=3x+6 -x=6 x=6*-1 x=-6

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