Question 696353
Let {{{ a }}} = cm3 of gold in jewelers 1 cm^3
Let {{{ b }}} = cm3 of silver in jewelers 1 cm^3
(1) {{{ 19a + 8b = 18 }}}
Note that in terms of units, this is
( gm / cm3 )x( cm3 ) + ( gm / cm3 ) = ( gm )
So I have ( gm ) = ( gm )
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Also,
(2) {{{ a + b = 1 }}}
( cm3 ) + ( cm3 ) = ( cm3 )
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Multiply both sides of (2) by {{{ 8 }}}
and subtract (2) from (1)
(1) {{{ 19a + 8b = 18 }}}
(2) {{{ -8a - 8b = -8 }}}
{{{ 11a = 10 }}}
{{{ a = 10/11 }}}
and, since
(2) {{{ a + b = 1 }}}
(2) {{{ b = 1/11 }}}
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There is 10/11 cm3 of gold in jewelers 1 cm^3
There is 1/11 cm3 of silver in jewelers 1 cm^3
The weight of gold in 10/11 cm3 is 
{{{ 19*(10/11) = 190/11 }}}
The weight of silver in 1/11 cm3 is
{{{ 8*(1/11) = 8/11 }}}
( weight of gold ) / ( weight of silver ) = {{{ ( 190 / 11 ) / ( 8 / 11 ) }}}
 {{{ ( 190 / 11 ) / ( 8 / 11 ) = 190/8 }}}
{{{ 23.75 }}}
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( weight of gold ) / ( weight of silver ) = 23.75
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check answer:
{{{ ( 19a ) / ( 8b ) = 23.75 }}}
{{{ a = 1 - b }}}
{{{ ( 19*( 1 - b ) ) / ( 8b ) = 23.75 }}}
{{{ ( 19 - 19b ) / ( 8b ) = 23.75 }}}
{{{ 19 - 19b = 190b }}}
{{{ 209b = 19 }}}
{{{ b = 19/209 }}}
{{{ b = 1/11 }}}
OK