How do you solve #\sqrt { 6x + 13} = 7#?

Question

624 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-27T20:48:20

See a solution process below:

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

#(sqrt(6x + 13))^2 = 7^2#

#6x + 13 = 49#

Next, subtract #color(red)(13)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#6x + 13 - color(red)(13) = 49 - color(red)(13)#

#6x + 0 = 36#

#6x = 36#

Now, divide each side of the equation by #color(red)(6)# to solve for #x# while keeping the equation balanced:

#(6x)/color(red)(6) = 36/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 6#

#x = 6#