Answer:
5 cats, 1 fish and 94 birds
Explanation:
Let
- c be the number of cats;
- b be the number of birds;
- f be the number of fish.
A kid bought 100 toy animals, then
![c+b+f=100](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1uwabf3ar1cfrb4rvpy22skbmq97vx5kko.png)
If a cat costs $10.00, then c cats cost $10c.
If a fish costs $3.00, then f fish cost $3f.
If a bird costs $0.50, then b birds cost $0.5b.
In total,
![10c+3f+0.5b=100](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ropxkhc0lgpmbc3geelhzlp1hsifatypt0.png)
We get the system of two equations:
![\left\{\begin{array}{l}c+b+f=100\\10c+3f+0.5b=100\end{array}\right.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6tcqf3fgkcbd2k73pe34vnqitrny6zyayb.png)
From the first equation:
![b=100-c-f](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ivqlmokvm3r095wsacorh075vooeoahdco.png)
Substitute it into the second equation:
![10c+3f+0.5(100-c-f)=100\\ \\10c+3f+50-0.5c-0.5f=100\\ \\9.5c+2.5f=50\\ \\95c+25f=500\\ \\19c+5f=100\\ \\f=20-(19)/(5)c](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4fx5rni6rnz26m0wsu6osrc9lsu3nfbxij.png)
Number f must be a whole number greater than 0, so
![20-(19)/(5)c>0\\ \\c<(100)/(19)\\ \\c\le 5(5)/(19)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/znxbzx6pagoitcvu78m75fkssnkq5hw0hs.png)
Number c must be a multiple of 5, thus the only possible value for c is 5.
When c = 5, then
![f=20-5\cdot (19)/(5)=20-19=1\\ \\b=100-5-1=94](https://img.qammunity.org/2020/formulas/mathematics/middle-school/smo4obgb1n10q2x2jv1gv1adq1o6cdzuug.png)