Question 891000
Let {{{ d[b] }}} = the number of dimes Barry has
Let {{{ q[b] }}} = the number of quarters Barry has
Let {{{ n[a] }}} = the number of nickels Andy has
Let {{{ d[a] }}} = the number of dimes Andy has
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(1) {{{ 10d[b] + 25q[b] = 550 }}} ( in cents )
(2) {{{ 10d[a] + 5n[a] = 240 }}}
(3) {{{ n[b] = n[a] }}}
(4) {{{ q[b] = d[a] }}}
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There are 4 equations and 4 unknowns,
so it's solvable
Substitute (4) into (2)
(2) {{{ 10d[a] + 5n[a] = 240 }}}
(2) {{{ 10q[b] + 5n[a] = 240 }}}
Substitute (3) into (2)
(2) {{{ 10q[b] + 5n[a] = 240 }}}
(2) {{{ 10q[b] + 5n[b] = 240 }}}
(2) {{{ 5n[b] + 10q[b] = 240 }}}
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Multiply both sides of (2) by {{{ 2 }}}
and subtract (2) from (1)
(1) {{{ 10d[b] + 25q[b] = 550 }}} 
(2) {{{ -10d[b] - 20q[b] = -480 }}}
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{{{ 5q[b] = 70 }}}
{{{ q[b] = 14 }}}
and, since
(4) {{{ q[b] = d[a] }}}
(4) {{{ d[a] = 14 }}}
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Back to (1)
(1) {{{ 10d[b] + 25q[b] = 550 }}} 
(1) {{{ 10d[b] + 25*14 = 550 }}} 
(1) {{{ 10d[b] = 550 - 350 }}}
(1) {{{ 10d[b] = 200 }}}
(1) {{{ d[b] = 20 }}}
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Back to (2)
(2) {{{ 10d[a] + 5n[a] = 240 }}}
(2) {{{ 10*14 + 5n[a] = 240 }}}
(2) {{{ 5n[a] = 240 - 140 }}}
(2) {{{ n[a] = 100/5 }}}
(2) {{{ n[a] = 20 }}}
Andy has 20 nickels and 14 dimes
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check:
(1) {{{ 10d[b] + 25q[b] = 550 }}} 
(1) {{{ 10*20 + 25*14 = 550 }}} 
(1) {{{ 200 + 350 = 550 }}}
(1) {{{ 550 = 550 }}}
OK
You can check the other equations