How do you find all real number solutions to #root3(6y+2)-2=0#?

1 Answer
Nov 16, 2016

Answer:

#y=6#

Explanation:

your equation#\rArr root3(6y+2)-2=0#

  • add 2 to both sides
    #root3(6y+2)=2#
  • cube both sides to get rid of cube-root
    #6y+2=2^3#
  • simplify the cube
    #6y+2=8#
  • subtract 2 from both sides
    #6y=8-2#
  • simplify
    #6y=6#
  • divide both sides by 6 to isolate #y# on left side
    #y=6#