Question 1020279
1.{{{N+D+Q=28}}}
2.{{{3N+Q=D}}}
3.{{{5N+10D+25Q=280}}}
Use eq. 2 to substitute for D in 1 and 3.
{{{N+3N+Q+Q=28}}}
{{{4N+2Q=28}}}
4.{{{2N+Q=14}}}
.
.
{{{5N+10(3N+Q)+25Q=280}}}
{{{5N+30N+10Q+25Q=280}}}
{{{35N+35Q=280}}}
5.{{{N+Q=8}}}
Now subtract 5 from 4 to eliminate Q.
{{{2N+Q-N-Q=14-8}}}
{{{N=6}}}
Then,
{{{2(6)+Q=14}}}
{{{12+Q=14}}}
{{{Q=2}}}
and
{{{3(6)+2=D}}}
{{{D=20}}}