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Mathematics 15 Online
OpenStudy (anonymous):

Solve for x in the proportion 9/27x = x/x+3 I understand : 9/(27x) = x/(x + 3) 1/(3x) = x/(x + 3) Cross multiply : x + 3 = 3x^2 3x^2 - x - 3 = 0

OpenStudy (anonymous):

looks fine

OpenStudy (anonymous):

x ≈ −0.85 and x ≈ 1.18<--- isn't this my answer

OpenStudy (anonymous):

?

OpenStudy (anonymous):

9/27x = x/(x + 3) 27x^2 - 9x - 27 = 0 3x^2 - x - 3 = 0 x = (1 - sqrt 37)/6 = -0.847 x = (1 + sqrt 37)/6 = 1.18

OpenStudy (anonymous):

Is it \[\frac{9}{27}x\]or,\[\frac{9}{27x}\]?

OpenStudy (anonymous):

same question with right hand side. x/(x+3) ?

OpenStudy (anonymous):

cross multiply: \[9(x+3)=27x(x)\]so,\[9x+27=27x^2\]or,\[3x^2-x-3=0\]So, yeah, assuming I have your equation correct I agree with your answer.

OpenStudy (anonymous):

thank you! ^^

OpenStudy (anonymous):

no prob :)

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