Kkathry922idn
Answered

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On a farm, there are 20 animals consisting of ducks and pigs. these are 52 animal legs. how many of each kind are there?
A. d=11;p=9
B. d=12;p=8
C. d=13;p=7
D. d=14;p=6

Sagot :

KAntoineDoixidn

✒️ EQUATION SYSTEM

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad \Large \:\: \rm D. \; d = 14 \:;\: p = 6 [/tex]

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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]

Let d and p be the number of ducks and pigs, respectively. Create two equations based on the given.

  • [tex] \begin{cases} d + p = 20 \\ 2d + 4p = 52 \end{cases} \quad \begin{align} \tt{(eq. \: 1)} \\ \tt{(eq. \: 2)} \end{align} [/tex]

Find either d or p in the first equation then substitute it to the second equation.

  • [tex] \begin{cases} d = 20 - p \\ 2d + 4p = 52 \end{cases} [/tex]

  • [tex] \begin{cases} d = 20 - p \\ 2(20-p) + 4p = 52 \end{cases} [/tex]

  • [tex] \begin{cases} d = 20 - p \\ 40 - 2p + 4p = 52 \end{cases} [/tex]

  • [tex] \begin{cases} d = 20 - p \\ \text - 2p + 4p = 52 - 40 \end{cases} [/tex]

  • [tex] \begin{cases} d = 20 - p \\ \text 2p = 12\end{cases} [/tex]

  • [tex] \begin{cases} d = 20 - p \\ \text 2p/2 = 12/2 \end{cases} [/tex]

  • [tex] \begin{cases} d = 20 - p \\ \text p = 6\end{cases} [/tex]

Thus, the value of p is 6. Substitute it to the first equation to find the value of d.

  • [tex] \begin{cases} d = 20 - 6 \\ \text p = 6\end{cases} [/tex]

  • [tex] \begin{cases} d = 14 \\ \text p = 6\end{cases} [/tex]

Therefore, there were 14 ducks and 6 pigs in the farm.

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