How do you solve #sqrt((5y)/6)-10=4# and check your solution?

2 Answers
May 31, 2017

Answer:

See a solution process below:

Explanation:

First, add #color(red)(10)# to each side of the equation to isolate the radical while keeping the equation balanced:

#sqrt((5y)/6) - 10 + color(red)(10) = 4 + color(red)(10)#

#sqrt((5y)/6) - 0 = 14#

#sqrt((5y)/6) = 14#

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

#(sqrt((5y)/6))^2 = 14^2#

#(5y)/6 = 196#

Now, multiply each side of the equation by #color(red)(6)/color(blue)(5)# to solve for #y# while keeping the equation balanced:

#color(red)(6)/color(blue)(5) xx (5y)/6 = color(red)(6)/color(blue)(5) xx 196#

#cancel(color(red)(6))/cancel(color(blue)(5)) xx (color(blue)(cancel(color(black)(5)))y)/color(red)(cancel(color(black)(6))) = 1176/5#

#y = 1176/5#

To check the solution we need to substitute #color(red)(1176/5)# for #color(red)(y)#, calculate each side of the equation and ensure the two results are equal:

#sqrt((5color(red)(y))/6) - 10 = 4# becomes:

#sqrt((5 xx color(red)(1176/5))/6) - 10 = 4#

#sqrt((color(red)(cancel(color(black)(5))) xx color(red)(1176/color(black)(cancel(color(red)(5)))))/6) - 10 = 4#

#sqrt(1176/6) - 10 = 4#

#+-sqrt(196) - 10 = 4#

#+-14 - 10 = 4#

#4 = 4# or #-24 = 4#

The #-14# result of the square root is extraneous.

Therefore, #4 = 4# and the solution is shown to be correct.

May 31, 2017

Answer:

#y = 235.2#

Explanation:

#sqrt((5y)/6) - 10 = 4#

#sqrt((5y)/6) = 4 + 10#

#sqrt((5y)/6) = 14#

#(5y)/6 = 14^2#

#(5y)/6 = 196#

#5y = 196 xx 6#

#5y = 1176#

#y = 1176 รท 5#

#color(blue)(y = 235.2#

We can now substitute #y# for #235.2# to prove our answer.

#sqrt((5y)/6) - 10 = 4#

#sqrt((5 xx 235.2)/6) - 10 = 4#

#sqrt(1176/6) - 10 = 4#

#sqrt(196) - 10 = 4#

#14 - 10 = 4#