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mixas84 [53]
3 years ago
8

There are thirteen animals in a barn. Some are chickens and some are pigs there are 40 legs in all. How many of each animal are

there
Mathematics
1 answer:
slega [8]3 years ago
3 0
Hi. In order to solve this word problem, you must make two equations and solve them together. To make it easy, we will let the letter "p" represent the pigs and the letter "c" for the chickens. So our first equation will be:

p + c = 13

Now to count the legs - we know that pigs have 4 legs and chickens have 2 so our next equation will be:

2c + 4p = 40

Using the first equation we see that c = 13 - p Now we insert this into the second equation and solve:

2x(13-p) +4 pm = 40
26 - 2p + 4p = 40
2p = 14
p = 7 (there are 7 pigs so now we know there are 6 chickens.

Answer: 7 pigs and 6 chickens

I hope this helps.

Take care,
Diana
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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
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Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

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Step-by-step explanation:

We have the following system of equations

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The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

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\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

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\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

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Step-by-step explanation:

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