Question 1153556
Original number of friends = x
<pr>
{{{525/x}}} = {{{(525/(x+6)) + 10}}}
<pr>
{{{525/x}}} = {{{(525/(x+6)) + (10(x+6)/(x+6))}}}
<pr>
{{{525/x}}} = {{{(525/(x+6)) + ((10x+60)/(x+6))}}}
<pr>
{{{525/x}}} = {{{(10x+585)/(x+6)}}}
<pr>
525(x+6) = x(10x+585)
<pr>
525x + 3150 = 10x² + 585x
<pr>
10x² + 60x - 3150 = 0
<pr>
x² + 6x - 315 = 0
<pr>
(x - 15)(x + 21) = 0
<pr>
x = 15 and x = -21.  Since x (the original number of friends) cannot be a negative number, this leaves us with x = 15.
<pr>
Original number of friends = x = 15