Question 1052880
Let {{{ n }}} = the original number of pocketbooks
{{{ 720/n }}} = the original price/pocketbook
{{{ 720/n - 6 }}} = the original price/pocketbook
reduced by 6 pesos/pocketbook
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[ reduced cost/pocketbook ]x[ new number of pocketbooks ] = [ same price ]
{{{ ( 720/n - 6 )*( n + 6 ) = 720 }}}
{{{ 720 - 6n + 4320/n - 36 = 720 }}}
{{{ -6n + 4320/n - 36 = 0 }}}
Multiply both sides by {{{ n }}}
{{{ -6n^2 + 4320 - 36n = 0 }}}
{{{ 6n^2 + 36n - 4320 = 0 }}}
{{{ n^2 + 6n - 720 = 0 }}}
Use quadratic formula
{{{ n = ( -b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 1 }}}
{{{ b = 6 }}}
{{{ c = -720 }}}
{{{ n = ( -6 +- sqrt( 6^2 - 4*1*(-720) ))/(2*1) }}}
{{{ n = ( -6 +- sqrt( 36 + 2880 )) / 2 }}}
{{{ n = ( -6 + sqrt( 2916 )) / 2 }}}
{{{ n = ( -6 + 54 )/2 }}}
{{{ n = 48/2 }}}
{{{ n = 24 }}}
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The original number of pocketbooks 
bought was 24
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{{{ 720/n = 720/24 }}}
{{{ 720/n = 30 }}}
The price/pocketbook was 30 pesos
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check answer:
{{{ ( 720/n - 6 )*( n + 6 ) = 720 }}}
{{{ ( 720/24 - 6 )*( 24 + 6 ) = 720 }}}
{{{ ( 30 - 6 )*30 = 720 }}}
{{{ 24*30 = 720 }}}
{{{ 720 = 720 }}}
OK