Question 850786
{{{ p }}} = number of pennies
{{{ d }}} = number of dimes
{{{ q }}} = number of quarters
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(1) {{{ p + d + q = 28 }}}
(2) {{{ p = d }}}
(3) {{{ 1*p + 10d + 25q = 193 }}} ( in cents )
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There are 3 unknowns and 3 equations,
so it's solvable
Multiply both sides of (1) by {{{ 25 }}} and
subtract (3) from (1)
(1) {{{ 25p + 25d + 25q = 700 }}}
(3) {{{ -p - 10d - 25q = -193 }}}
{{{ 24p + 15d = 507 }}}
and, since {{{ p = d }}},
{{{ 24d + 15d = 507 }}}
{{{ 39d = 507 }}}
{{{ d = 13 }}}
{{{ p = 13 }}}
and
(1) {{{ p + d + q = 28 }}}
(1) {{{ 13 + 13 + q = 28 }}}
(1) {{{ 26 + q = 28 }}}
(1) {{{ q = 2 }}}
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He has:
13 pennies
13 dimes
2 quarters
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check:
(3) {{{ 1*p + 10d + 25q = 193 }}}
(3) {{{ 1*13 + 10*13 + 25*2 = 193 }}}
(3) {{{ 13 + 130 + 50 = 193 }}}
(3) {{{ 193 = 193 }}}
OK