How do you solve #3/7x+1/2=3# by clearing the fractions?

#3/7x+1/2=3#

#3/7x=3-1/2# isolate terms with variable
#3/7x=\color(crimson)(6/2)-1/2# common denominator on right side
#3/7x=5/2# simplify

#3/7x(\color(teal)(2/2))=5/2(\color(teal)(7/7))# multiply to get common denominator overall
#\color(seagreen)(6/14)x=\color(seagreen)(35/14)#

#x=\color(indigo)(35/14*14/6)# divide (multiply by reciprocal for fractions) to get variable
#x=35/6#

Symbolab step-by-step

#3/7x+1/2=3#

#:.(6x+7)/14=42/14#

multiply both sides by #14#

#:.6x+7=42#

#:.6x=42-7#

#:.6x=35#

#:.color(blue)(x=35/6# or #color(blue)( 5 5/6#

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check:-

#:.3/7(color(blue)(35/6))+1/2=3#

#:.cancel3/cancel7^color(blue)1xxcancel35^color(blue)5/cancel6^color(blue)2+1/2=3#

#:.5/2+1/2=3#

#6/2=3#

#:.color(blue)(3=3#

When you have an EQUATION with fractions, you can get rid of the denominators immediately by multiplying through by the LCM of the denominators to cancel them.

#3/7x+1/2=3" "(LCD =color(blue)(14))#

#(color(blue)(14)xx3x)/7+(color(blue)(14)xx1)/2=color(blue)(14)xx3" "(larrxx color(blue)(14))#

#(color(blue)(cancel14^2)xx3x)/cancel7+(color(blue)(cancel14^7)xx1)/cancel2=color(blue)(14)xx3" "larr# cancel denominators

#6x+7 =42" "larr# no fractions!

#6x = 35#

#x = 35/6#

#x = 5 5/6#