How do you find all real number solutions to #root3(6y+2)-2=0#?
1 Answer
Nov 16, 2016
Explanation:
your equation
- 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#
