How do you solve #2root3(7b-1)-4=0#?

1 Answer
Jun 27, 2017

Answer:

See a solution process below:

Explanation:

First, add #color(red)(4)# to each side of the equation to isolate the radical term while keeping the equation balanced:

#2root(3)(7b - 1) - 4 + color(red)(4) = 0 + color(red)(4)#

#2root(3)(7b - 1) - 0 = 4#

#2root(3)(7b - 1) = 4#

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

#(2root(3)(7b - 1))/color(red)(2) = 4/color(red)(2)#

#(color(red)(cancel(color(black)(2)))root(3)(7b - 1))/cancel(color(red)(2)) = 2#

#root(3)(7b - 1) = 2#

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

#(root(3)(7b - 1))^color(red)(3) = 2^color(red)(3)#

#7b - 1 = 8#

Next, add #color(red)(1)# to each side of the equation to isolate the #b# term while keeping the equation balanced:

#7b - 1 + color(red)(1) = 8 + color(red)(1)#

#7b - 0 = 9#

#7b = 9#

Now, divide each side of the equation by #color(red)(7)# to solve for #b# while keeping the equation balanced:

#(7b)/color(red)(7) = 9/color(red)(7)#

#(color(red)(cancel(color(black)(7)))b)/cancel(color(red)(7)) = 9/7#

#b = 9/7#