Question 1050005
Let {{{ n }}} = the number of nickels
Let {{{ d }}} = the number of dimes
Let {{{ q }}} = the number of quarters
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(1) {{{ n + d + q = 35 }}}
(2) {{{ .05n + .1d + .25q = 6 }}}
(3) {{{ n = d + 5 }}}
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(2) {{{ 5n + 10d + 25q = 600 }}}
(2) {{{ n + 2d + 5q = 120 }}}
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(2) {{{ d + 5 + 2d + 5q = 120 }}}
(2) {{{ 3d + 5q = 115 }}}
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(1) {{{ d + 5 + d + q = 35 }}}
(1) {{{ 2d + q = 30 }}}
Multiply this result by {{{ 5 }}}
and subtract (2) from (1)
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(1) {{{ 10d + 5q = 150 }}}
(2) {{{ -3d - 5q = -115 }}}
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{{{ 7d = 35 }}}
{{{ d = 5 }}}
and
(3) {{{ n = d + 5 }}}
(3) {{{ n = 5 + 5 }}}
(3) {{{ n = 10 }}}
and
(1) {{{ n + d + q = 35 }}}
(1) {{{ 10 + 5 + q = 35 }}}
(1) {{{ q = 20 }}}
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There are 10 nickels, 5 dimes, 20 quarters
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(2) {{{ .05n + .1d + .25q = 6 }}}
(2) {{{ .05*10 + .1*5 + .25*20 = 6 }}}
(2) {{{ .5 + .5 + 5 = 6 }}}
(2) {{{ 6 = 6 }}}
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I like to use a different variable for each
type of coin. All through the calculations,
I am sure of what I am looking at. Then, at the 
end I can put the results all in terms of either
{{{n}}}, {{{d}}}, or {{{q}}}