Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
$z + z + 3 + z/2 = 23$, 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
4  
nicest granma everrr !! –  FromCuba Nov 3 '13 at 5:03

2 Answers 2

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, $11+8+4=23$.

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.