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Sagot :
[tex]x-\ amount\ of\ catfish\\\\
y-\ amount\ of\ blue\ gill\\\\\
\left \{ {{x+y=75} \atop {2y+3x=175}} \right. |using\ substitution\ method:\ x=75-y\\\\
2y+3(75-y)=175\\
2y+225-3y=175\\
-y=-50\ \ |*(-1)
y=50\\
x=75-50=25\\
there\ were\ 25\ catfish\ put\ in\ the\ pond[/tex]
You have to work this out through a process of trial and error. Firstly, let's start simple, by dividing the 175 by 5. 5 because if the number of blue gill is doubled and catfish tripled, there will be five groups rather than two, three of the groups being identical, and the other two also being identical.
175/5 = 35
35 x 2 = 70 x
As you can see, this hasn't worked, as the original amount ends up being 70 rather than 75. Now try again. Split the 175 up into two numbers, one that is divisible by three, the other that is divisible by two. For example, 100 and 75.
100/2 = 50
75/3 = 25
50 + 25 = 75
This way worked, though usually you might have to go on guessing for quite a while. The original amount of catfish in the pond would be 25 (because the number of catfish was tripled and the 75 was divided by three).
So yeah, the answer's 25. Hope that helped. Let me know if there's anything I need to explain better ;)
175/5 = 35
35 x 2 = 70 x
As you can see, this hasn't worked, as the original amount ends up being 70 rather than 75. Now try again. Split the 175 up into two numbers, one that is divisible by three, the other that is divisible by two. For example, 100 and 75.
100/2 = 50
75/3 = 25
50 + 25 = 75
This way worked, though usually you might have to go on guessing for quite a while. The original amount of catfish in the pond would be 25 (because the number of catfish was tripled and the 75 was divided by three).
So yeah, the answer's 25. Hope that helped. Let me know if there's anything I need to explain better ;)
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