Question 963081
given: area=540sq.yd ; {{{ L=(1 & 2/3)*w }}}

{{{ A=L*W }}}

substitute values
{{{ 540=(1 & 2/3)*w*w }}}
{{{ 540=(1 & 2/3)w^2 }}}

i changed the mixed fraction to improper so that it would be easy to multiply and divide
{{{ 540=(5/3)w^2 }}}

multiply both sides by 3 so that 3 in the denominator would be eliminated
{{{ 3(540)=((5/3)w^2)3 }}}
{{{ 1620=5w^2 }}}

divide both sides by 5 so that 5 would be eliminated and w^2 would remain
{{{ 1620/5=(5w^2)/5 }}}
{{{ 324=w^2 }}}

square both sides so that only w would remain
{{{ sqrt(324)=sqrt(w^2) }}}
{{{ 18=w }}}

we already have the width=18
substitute to formula
{{{ A=L*W }}}
{{{ 540=L*18 }}}
{{{ 540=18L }}}
{{{ 540/18=(18L)/18 }}}
{{{ 30=L }}}

therefore 30yards x 18yards is the dimension of the swimming pool