Question 834630
Let,
1.{{{D=P-4}}}
2.{{{D=1+2Q}}} 
3.{{{N=Q+1}}} 
{{{0.01P+0.05N+0.10D+0.25Q=1.76}}}
4.{{{P+5N+10D+25Q=176}}}
From eq. 1,
{{{P=D+4}}}
From eq. 2,
{{{2Q=D-1}}}
{{{Q=(1/2)(D-1)=D/2-1/2}}}
from eq. 3,
{{{N=Q+1=D/2-1/2+1=D/2+1/2}}}
Now substitute all of those into eq. 4 and solve for D.
{{{(D+4)+5(D/2+1/2)+10D+25(D/2-1/2)=176}}}
Multiply by 2 to get rid of fractions,
{{{2(D+4)+5(D+1)+20D+25(D-1)=352}}}
{{{2D+8+5D+5+20D+25D-25=352}}}
{{{52D=352+25-5-8}}}
{{{52D=364}}}
{{{D=7}}}
Then back substituting,
{{{P=D+4=7+4=11}}}
{{{N=D/2+1/2=7/2+1/2=8/2=4}}}
{{{Q=D/2-1/2=7/2-1/2=6/2=3}}}