Question 389785
Let {{{x}}} = the width of the walk
The area of the swimming pool is {{{6*10 = 60}}} m2
The walk has the same area
Therefore the area of the pool plus the area of the walk
is {{{2*60 = 120}}} m2
Each side of the pool is increased by {{{2x}}}
{{{(6 + 2x)*(10 + 2x) = 120}}}
{{{60 + 20x + 12x + 4x^2 = 120}}}
{{{4x^2 + 32x - 60 = 0}}}
{{{x^2 + 8x - 15 = 0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{a = 1}}}
{{{b = 8}}}
{{{c = -15}}}
{{{x = (-8 +- sqrt( 8^2-4*1*(-15) ))/(2*1) }}} 
{{{x = (-8 +- sqrt( 64 + 60 ))/2 }}}
{{{x = (-8 + sqrt(124))/2}}}
{{{x =( -8 + 11.136)/2}}}
{{{x = 3.136/2}}}
{{{x = 1.568}}}
The width of the walk is 1.568 m
check answer:
{{{(6 + 2x)*(10 + 2x) = 120}}}
{{{(6 + 3.136)*(10 + 3.136) = 120}}}
{{{9.136*13.136 = 120}}}
{{{12.272*16.272 = 120}}}
{{{120.010 = 120}}}
close enough