anonymous
  • anonymous
help please Addison and Caden prefer tetra fish to goldfish. Addison buys 5 neon tetras and 4 cardinal tetras for $15.50. Caden buys 6 neon tetras and 10 cardinal tetras for $26.40. A. Write a system of equations that can be used to find n, the cost of 1 neon tetra, and c, the cost of 1 cardinal tetra. B.What is the cost of 1 neon tetra?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Owlcoffee
  • Owlcoffee
Here, you have to set up a system of equalities that allows us to model the situation. Let's first model what Addison bought; this being 5 neon tetras (n) and 4 cardinal tetras (c) and paid 15.5$ so therefore, a model would be: \[5n +4c=15,50\] That will be the first equation, now for Cadens model; she bought 6 neon tetras and 10 cardinal tetra, and paid 26.40$ which means that the model for her purchse is: \[6n +10c=26,40\] Now, in order to find the prices for each type of fish we will then build the system of equation, and it will be a 2x2 type, since we have two variables and two equations: \[5n+4c=15,50\] \[6n+10c=26,40\]
Owlcoffee
  • Owlcoffee
Now you can solve that using any method you have been taught.
anonymous
  • anonymous
thank youuuu

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More answers

anonymous
  • anonymous
yay @Tommynaiter
anonymous
  • anonymous
You need to setup 2 equations with 2 variables. You get the following: \[5n+4c=15.50\]\[6n+10c=26.40\] Here is n neon and c is cardinal. Do you see where the equations come from?
anonymous
  • anonymous
yes but i have trouble finding the price
anonymous
  • anonymous
Alright. So do you know the method substitution?
anonymous
  • anonymous
yes
anonymous
  • anonymous
So you want to use that.
anonymous
  • anonymous
sure
anonymous
  • anonymous
So lets do it in small steps. Start by isolating n in the first equation. \[5n+4c=15.50\]\[n=?\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
Do you know how to isolate n?
anonymous
  • anonymous
you divide
anonymous
  • anonymous
First we subtract 4c. \[5n+4c=15.50\]\[5n=15.50-4c\] Now we divide with 5 \[n=3.1-\frac{4}{5}c\]
anonymous
  • anonymous
subtract 4c
anonymous
  • anonymous
Yeap, it looks like you know how to solve n.
anonymous
  • anonymous
So the next step is to insert the value for n into the 2nd equation.
anonymous
  • anonymous
ok
anonymous
  • anonymous
We had the 2nd equation: \[6n+10c=26.40\] Insert the values for n: \[6*(3.1-\frac{4}{5}c)+10c=26.40\]
anonymous
  • anonymous
Do you follow what I just did?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Good, so now we want to isolate c. Then we have the price for c.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Can you do that?
anonymous
  • anonymous
you divide
anonymous
  • anonymous
right
anonymous
  • anonymous
There is a couple of steps. First I would get rid of the parentheses.
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[6∗(3.1−\frac{4}{5}c)+10c=26.40\]\[6*3.1=18.6\]\[6*(-\frac{4}{5}c)=-\frac{24}{5}c\] So you get \[18.6-\frac{24}{5}c+10c=26.40\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
So we continue to solve c. Subtract 18.6 \[-\frac{24}{5}c+10c=7.8\] add the c's together \[\frac{26}{5}c=7.8\] multiply with 5 \[26c=39\] divide with 26 \[c=1.5\]
anonymous
  • anonymous
Do you understand how I solved c?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Good, so now that we know the price for cardinal. We just insert the price into either one of the equations and find a value for n. \[5n+4c=15.50\] insert \[c=1.5\] \[5n+4*1.5=15.50\]\[5n+6=15.50\] subtract 6 \[5n=9.50\] divide with 5 \[n=1.90\]
anonymous
  • anonymous
So the price for the cardinal (c) is 1.5 and the price for neon (n) is 1.9
anonymous
  • anonymous
ok thank youuuu
anonymous
  • anonymous
Yea, I hope you understand it :) else just ask
anonymous
  • anonymous
I am always willing to help, if you'd like you can follow me to see if im online :)
anonymous
  • anonymous
ok thanks :)
anonymous
  • anonymous
You're welcome, have a good day @ameliagrace417 :D
anonymous
  • anonymous
you tooo @Tommynaiter :)

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