How do you solve #\sqrt { 12t - 5} = \sqrt { 9t + 14}#?

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Answer 1 · 2017-05-30T16:46:20

See a solution process below:

First, square each side of the equation to eliminate the radicals while keeping the equation balanced:

#(sqrt(12t - 5))^2 = (sqrt(9t + 14))^2#

12t - 5 = 9t + 14#

Next add #color(red)(5)# and subtract #color(blue)(9t)# from each side of the equation to isolate the #t# term while keeping the equation balanced:

#-color(blue)(9t) + 12t - 5 + color(red)(5) = -color(blue)(9t) + 9t + 14 + color(red)(5)#

#(-color(blue)(9) + 12)t - 0 = 0 + 19#

#3t = 19#

Now, divide each side of the equation by #color(red)(3)# to solve for #t# while keeping the equation balanced:

#(3t)/color(red)(3) = 19/color(red)(3)#

#(color(red)(cancel(color(black)(3)))t)/cancel(color(red)(3)) = 19/3#

#t = 19/3#

Answer 2 · 2017-06-12T11:41:10

#color(blue)(t=19/3#

#sqrt(12t-5)=sqrt(9t+14)#

square both sides

#:.(sqrt(12t-5))^2=(sqrt(9t+14))^2#

#:.sqrta xx sqrta=a#

#:.sqrt(12t-5) xx sqrt(12t-5)=sqrt(9t+14) xx sqrt(9t+14)#

#:.12t-5=9t+14#

#:.12t-9t=14+5#

#:.3t=19#

#:.color(blue)(t=19/3#

Substitute#color(blue)(t=19/3#

#:.sqrt(12(19/3)-5)=sqrt(9(19/3)+14)#

#:.sqrt((228/3)-5)=sqrt((171/3)+14)#

#:.sqrt(76-5)=sqrt(57+14)#

#:.color(blue)(sqrt71=sqrt71#