Let there be [tex]h[/tex] 50 cent coins (call them halfs), [tex]d[/tex] 10 cent coins (dimes) and [tex]n[/tex] 5 cent coins (nickles).
81 coins:
[tex]h+d+n=81[/tex]
Three more dimes than halfs:
[tex]d=h+3[/tex]
Nickels are twice halfs plus dimes:
[tex]n=2(h+d)[/tex]
Three equations in three unknowns. We'll eliminate [tex]n[/tex] first:
[tex]n=81-(h+d)=2(h+d)[/tex]
[tex]81=3(h+d)[/tex]
[tex]h+d=27[/tex]
Substituting [tex]d=h+3[/tex]
[tex]h+h+3=27[/tex]
[tex]2h=24[/tex]
[tex]h=12[/tex]
[tex]d=h+3=15[/tex]
[tex]n=2(h+d)=2(27)=54[/tex]
Check we have the right number of coins:
[tex]h+d+n=12+15+54=81 \quad\checkmark[/tex]
Good. The amount of money is
[tex]50h+10d+5n=0.50(12)+0.10(15)+0.05(54)=\$10.20[/tex]
Part 2.
a for amount Tim gave cousin. Initially Tim has $146.65, cousin $10.55
After he gave the money Tim had twice as much:
[tex]146.65 - a = 2(10.55 + a)[/tex]
[tex]146.65 - a = 21.10 + 2a[/tex]
[tex]146.65 - 21.10 = 3a[/tex]
[tex]a = \dfrac{146.65 - 21.10}{3}= 41.85[/tex]
That's the answer to b. After the transfer the cousin has [tex]10.55+41.85=\$52.40[/tex]
