How do you solve #2sqrt(3x+5)+7=16#?

1 Answer
May 9, 2017

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(7)# from each side of the equation to isolate the radical term while keeping the equation balanced:

#2sqrt(3x + 5) + 7 - color(red)(7) = 16 - color(red)(7)#

#2sqrt(3x + 5) + 0 = 9#

#2sqrt(3x + 5) = 9#

Next, divide each side of the equation by #color(red)(2)# to isolate the radical while keeping the equation balanced:

#(2sqrt(3x + 5))/color(red)(2) = 9/color(red)(2)#

#(color(red)(cancel(color(black)(2)))sqrt(3x + 5))/cancel(color(red)(2)) = 9/2#

#sqrt(3x + 5) = 9/2#

Then, square each side of the equation to eliminate the radical while keeping the equation balanced:

#(sqrt(3x + 5))^2 = (9/2)^2#

#3x + 5 = 81/4#

Next, subtract #color(red)(5)# from each side of the equation to isolate the #x# term:

#3x + 5 - color(red)(5) = 81/4 - color(red)(5)#

#3x + 0 = 81/4 - (4/4 xx color(red)(5))#

#3x = 81/4 - 20/4#

#3x = 61/4#

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

#(3x)/color(red)(3) = (61/4)/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 61/12#

#x = 61/12#