Question 1134727

You have {{{50}}} coins, which total ${{{6.95}}}. They are all dimes and quarters.

let dimes be {{{d}}} and quarters {{{q}}}

{{{d +q=50}}}...solve or dimes
{{{d =50-q}}}........eq.1

total ${{{6.95}}},
dime is ${{{ 0.10}}}
quarter is ${{{ 0.25}}}

then we have

{{{ 0.10d+0.25q=6.95}}}.....substitute dime from eq.1

{{{ 0.10(50-q)+0.25q=6.95}}}.........solve for {{{q}}}

{{{ 5-0.10q+0.25q=6.95}}}

{{{ 0.15q=6.95-5}}}

{{{ 0.15q=1.95}}}

{{{ q=1.95/0.15}}}

{{{ q=13}}}

go to {{{d =50-q}}}........eq.1, substitute {{{q}}}

{{{d =50-13}}}

{{{d =37}}}

so, you have {{{37}}} dimes and {{{ 13}}} quarters 


check:
{{{ 0.10d+0.25q=6.95}}}
{{{ 0.10*37+0.25*13=6.95}}}
{{{ 3.7+3.25=6.95}}}
{{{ 6.95=6.95}}}