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Mathematics 21 Online
OpenStudy (anonymous):

solve for x 2x/7 + x/4 = 3

OpenStudy (anonymous):

\[Ax/B + Cx/D = (ADx + BCx)/BD = x(AD + BC)/BD\]

OpenStudy (anonymous):

boooooo! bad answer! solve 4 x!!!!!

OpenStudy (anonymous):

2x/7 + x/4 = 3 2x/7 * 4/4 + x/4 * 7/7 = 3 * get common denominators then proceed :)

OpenStudy (anonymous):

help any?

OpenStudy (anonymous):

\[x=3*BD/(AD+BC)\] There, x is solved for.

OpenStudy (anonymous):

2x/7 + x/4 = 3 | Original equation 8x/28 + 7x/28 = 84/28 | Make common denominators of 2x/7, x/4 and 3 15x/28 = 84/28 | Add 8x/28 and 7x/28 together x = 84/28*28/15 | Divide by 15/28 on both sides (multiply by reciprocal) x = 84/15 | Almost there x = 28/ 5 | Simplify the fraction by dividing 3 on both numerator and denominator And that's it! :)

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