Answer to Question #225915 in Chemical Engineering for Lokika

Question #225915

The value of 1/D²+3D (2 + 6x) =


A) x


B) x² C) 0 D) 1


1
Expert's answer
2021-08-19T15:23:02-0400

"D^2+3D\\left(2+6x\\right)=0\\\\\n\\mathrm{Subtract\\:}D^2\\mathrm{\\:from\\:both\\:sides}\\\\\nD^2+3D\\left(2+6x\\right)-D^2=0-D^2\\\\\n\\mathrm{Simplify}\\\\\n3D\\left(2+6x\\right)=-D^2\\\\\n\\mathrm{Divide\\:both\\:sides\\:by\\:}3D;\\quad \\:D\\ne \\:0\\\\\n\\frac{3D\\left(2+6x\\right)}{3D}=\\frac{-D^2}{3D};\\quad \\:D\\ne \\:0\\\\\n2+6x=-\\frac{D}{3};\\quad \\:D\\ne \\:0\\\\\n\\mathrm{Subtract\\:}2\\mathrm{\\:from\\:both\\:sides}\\\\\n2+6x-2=-\\frac{D}{3}-2;\\quad \\:D\\ne \\:0\\\\\n\\mathrm{Simplify}\\\\\n6x=-\\frac{D}{3}-2;\\quad \\:D\\ne \\:0\\\\\n\\mathrm{Divide\\:both\\:sides\\:by\\:}6;\\quad \\:D\\ne \\:0\\\\\n\\frac{6x}{6}=-\\frac{\\frac{D}{3}}{6}-\\frac{2}{6};\\quad \\:D\\ne \\:0\\\\\nx=\\frac{-D-6}{18};\\quad \\:D\\ne \\:0"


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