How do you solve #sqrt(7r+2)+3=7# and check your solution?

1 Answer
Aug 9, 2017

Answer:

See a solution process below:

Explanation:

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

#sqrt(7r + 2) + 3 - color(red)(3) = 7 - color(red)(3)#

#sqrt(7r + 2) + 0 = 4#

#sqrt(7r + 2) = 4#

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

#(sqrt(7r + 2))^2 = 4^2#

#7r + 2 = 16#

Then, subtract #color(red)(2)# from each side of the equation to isolate the #r# term while keeping the equation balanced:

#7r + 2 - color(red)(2) = 16 - color(red)(2)#

#7r + 0 = 14#

#7r = 14#

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

#(7r)/color(red)(7) = 14/color(red)(7)#

#(color(red)(cancel(color(black)(7)))r)/cancel(color(red)(7)) = 2#

#r = 2#

To validate the solution substitute #color(red)(2)# for #color(red)(r)# in the original equation and calculate the result to ensure both sides of the equation are equal (remember, the square root of a number produces a positive and negative result):

#+-sqrt(7color(red)(r) + 2) + 3 = 7# becomes:

#+-sqrt((7 * color(red)(2)) + 2) + 3 = 7#

#+-sqrt(14 + 2) + 3 = 7#

#+-sqrt(16) + 3 = 7#

#-4 + 3 = 7# and #4 + 3 = 7#

#-1 != 7# and #7 = 7#

The negative result of the radical is an extraneous solution.

The positive result shows the solution is correct.