Question 448512
Let {{{a}}} = pounds of $4 candy needed
Let {{{b}}} = pounds of $6 candy needed
Let {{{c}}} = pounds of $10 candy needed
given:
(1) {{{ a + b + c = 100 }}}
(2) {{{ 4a + 6b + 10c = 7.6*100 }}}
(2) {{{ 4a + 6b + 10c = 760 }}}
(2) {{{ 2a + 3b + 5c = 380 }}}
(3) {{{ c = a + b }}}
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There are 3 equations and 3 unknowns, so it's solvable
(3) {{{ c = a + b }}}
(3) {{{ a + b - c = 0 }}}
Add (3) and (1)
(1) {{{ a + b + c = 100 }}}
(3) {{{ a + b - c = 0 }}}
{{{ 2a + 2b = 100 }}}
{{{ a + b = 50 }}}
From (3), 
{{{ c = 50 }}}
Substitute this into (2)
(2) {{{ 2a + 3b + 5*50 = 380 }}}
(2) {{{ 2a + 3b + 250 = 380 }}}
(2) {{{ 2a + 3b = 130 }}}
Multiply both sides of (3) by {{{2}}}
and subtract (3) from (2)
(2) {{{ 2a + 3b = 130 }}}
(3) {{{ -2a - 2b = -100 }}}
{{{ b = 30 }}}
and, since
{{{ a+ b = 50 }}}
{{{ a + 30 = 50 }}}
{{{ a = 20 }}}
20 pounds of $4 candy are needed
30 pounds of $6 candy are needed
50 pounds of $10 candy are needed
check answer:
(2) {{{ 2a + 3b + 5c = 380 }}}
(2) {{{ 2*20 + 3*30 + 5*50 = 380 }}}
(2) {{{ 40 + 90 + 250 = 380 }}}
(2) {{{ 380 = 380 }}}
OK