Child's homework equation related to algebra

My grandson's homework.... There are 23 fish. Guppies (G) are 3 more than Z-fish (Z). There are 2 times the Z-fish as Goldfish (GF). How many of each? .... I can see there are 8 Z-fish (Z), 11 Guppies (G) & 4 Goldfish (GF).... HOWEVER, When I try to prove the equation it's wrong.. Z + (Z+3) +(Z/2) = 23... equation? So Z + (Z/2) = 20/2 (=10).... not correct..so.. answer should be 12 (if Z=8, G =11, GF=4 ). help? I don't know what I'm doing wrong. I can do a simple one if two unknowns but can't do this 3 unknowns with the fraction.

-
$z + z + 3 + z/2 = 23$, multiply everything by $2$, so we have $4z+6+z = 46$, so $5z = 40$ or $z = 8$. You are a great grandma! :-) –  Amzoti Nov 3 '13 at 5:01
nicest granma everrr !! –  ILoveMath Nov 3 '13 at 5:03

The initial system given by the problem is:

\left\{\begin{align*} &G+Z+GC=23\\ &G=Z+3\\ &Z=2GF \end{align*}\right.

The second and third equations make it easy to express $G$ and $GF$ in terms of $Z$: we already have $G=Z+3$, and from the third equation we get $GF=\frac12Z$. Substitute these into the first equation, and you’ll have an equation with just the single unknown $Z$:

$$(Z+3)+Z+\frac12Z=23\;.$$

I expect that both you and your grandson can handle that without much trouble, but I’ll go ahead and finish it off. Simplifying, we have

\begin{align*} &\frac52Z+3=23\;,\\\\ &\frac52Z=20\;,\text{ and}\\\\ &Z=\frac25\cdot20=8\;, \end{align*}

so $G=Z+3=11$ and $GF=\frac12Z=4$. And as a check, $11+8+4=23$.

-
Would the downvoter care to explain what’s wrong with the answer? –  Brian M. Scott Nov 3 '13 at 20:00

$$G+GF+Z=23\\G=Z+3\\Z=2GF\\G=Z+3=2GF+3\\2GF+3+GF+2GF=23\\5GF=20\\GF=20/5=4\\Z=2GF=2(4)=8\\G=2GF+3=2(4)+3=11\\\boxed{GF=4,Z=8,G=11} :)$$

-