<PRE><B>Question 181145</B>
I am assuming it is (1/3)x= x/3....not 1/(3x)

sqr(x-6) - sqr((1/3)x)=0
sqr(x-6) = sqr (1/3x)
(sqr(x-6))^2 = (sqr(1/3x))^2
x-6 = 1/3x
x-1/3x -6 = 1/3x - 1/3x
2/3x - 6 = 0
2/3x-6+6 = 0+6
2/3x = 6
3(2/3x) = 6(3)
2x = 18
2x/2 = 18/2
x = 9

check

sqr(9-6) - sqr(1/3 * 9)
sqr(3) - sqr (9/3)
sqr(3) - sqr (3)
0

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