How do you solve #\sqrt { 2x - 5} + 3= 11#?

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Answer 1 · 2017-07-16T21:14:15

See a solution process below:

First, subtract #color(red)(3)# from each side of the equation to isolate the radical while keeping the equation balanced:

#sqrt(2x - 5) + 3 - color(red)(3) = 11 - color(red)(3)#

#sqrt(2x - 5) + 0 = 8#

#sqrt(2x - 5) = 8#

Next, square both sides of the equation to eliminate the radical while keeping the equation balanced:

#(sqrt(2x - 5))^2 = 8^2#

#2x - 5 = 64#

Then, add #color(red)(5)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x - 5 + color(red)(5) = 64 + color(red)(5)#

#2x - 0 = 69#

#2x = 69#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = 69/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 69/2#

#x = 69/2#