Question 563364
If {{{x}}} = the width of the walkway, the 
length and width both increase by {{{ 2x }}}
given:
{{{ ( 10 + 2x )*( 20 + 2x ) = 10*20 + 136 }}}
{{{ 4*( 5 + x )*( 10 + x ) = 200 + 136 }}}
{{{ 4*( 50  + 10x + 5x + x^2 ) = 336 }}}
{{{ x^2 + 15x + 50 = 84 }}}
{{{ x^2 + 15x - 34 = 0 }}}
Use quadratic equation
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 1 }}}
{{{ b = 15 }}}
{{{ c = -34 }}}
{{{ x = (-15 +- sqrt( 15^2 - 4*1*(-34) ))/(2*1) }}} 
{{{ x = (-15 +- sqrt( 225 + 136 )) / 2 }}} 
{{{ x = (-15 +- sqrt( 361 )) / 2 }}} 
{{{ x = (-15 + 19) / 2 }}} 
{{{ x = 4/2 }}}
{{{ x = 2 }}} ( I can ignore the (-) square root )
The walkway is 2 ft wide
check:
{{{ ( 10 + 2*2 )*( 20 + 2*2 ) = 10*20 + 136 }}}
{{{ 14*24 = 336 }}}
{{{ 336 = 336 }}}
OK