How do you solve #\sqrt{x+15}=\sqrt{3x-3}#?

1 Answer
Mar 1, 2018

Answer:

#x=9#

Explanation:

#"To 'undo' the radicals square both sides"#

#(sqrt(x+15))^2=(sqrt(3x-3))^2#

#rArrx+15=3x-3#

#"subtract "(x+15)" from both sides"#

#rArr0=2x-18#

#rArr2x=18rArrx=9#

#color(blue)"As a check"#

#"substitute "x=9" into the equation and if both sides are"#
#"equal then it is the solution"#

#"left side "=sqrt24=2sqrt6#

#"right side "=sqrt24=2sqrt6#

#rArrx=9" is the solution"#