How do you solve #4sqrtx=sqrt( 9x+9)#?

First, square both sides:
#(4sqrtx)^color(blue)2 = (sqrt(9x+9))^color(blue)2#

#16x = 9x+9#

Subtract #color(blue)9x# from both sides of the equation:
#16x quadcolor(blue)(-quad9x) = 9x + 9 quadcolor(blue)(-quad9x)#

#7x = 9#

Finally, divide both sides by #7#:
#(7x)/color(blue)7 = 9/color(blue)7#

So the answer is:
#x = 9/7#

However, we should check our work by plugging our solution back into the original equation:
#4sqrt(9/7) = sqrt(9(9/7) + 9)#

#4*3/sqrt7 = sqrt(81/7 + 63/7)#

#12/sqrt7 = sqrt(144/7)#

#12/sqrt7 = 12/sqrt17#

Both sides of the equation equal, meaning that #x = 9/7# is true.

Hope this helps!

First of All,

#color(white)(xxx)4sqrt(x) = sqrt(9x + 9)#

#rArr 16x = 9x + 9# [Square both sides]

#rArr 16x - 9x = 9# [Transpose #9x# to L.H.S.]

#rArr 7x = 9#

#rArr x = 9/7#.

So, #x = 9/7#.

But if we substitute #x = 9/7# in the equation, we get,

L.H.S. = #4sqrt(9/7) = sqrt(16 * 9/7) = sqrt(144/7)#

R.H.S = #sqrt(9*9/7 + 9) = sqrt(81/7 + 9) = sqrt((81 + 63)/7) = sqrt(144/7)#

So, L.H.S. = R.H.S.

So #x = 9/7# is indeed a solution for this equation.

Hope this helps.