How do you solve #\sqrt { 11x - 10} = 12#?

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Answer 1 · 2017-07-16T01:49:15

See a solution process below:

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

#(sqrt(11x - 10))^2 = 12^2#

#11x - 10 = 144#

Next, add #color(red)(10)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#11x - 10 + color(red)(10) = 144 + color(red)(10)#

#11x - 0 = 154#

#11x = 154#

Now, divide each side of the equation by #color(red)(11)# to solve for #x# while keeping the equation balanced:

#(11x)/color(red)(11) = 154/color(red)(11)#

#(color(red)(cancel(color(black)(11)))x)/cancel(color(red)(11)) = 14#

#x = 14#