Respuesta :

DeanR
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]


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