Solve; \(\frac{P}{2} + \frac{k}{3}\) = 5 and 2p = k = 6 simultaneously
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Correct Answer: Option C
Explanation:
\(\frac{P}{2} + \frac{k}{3}\) = 5
\(\frac{3p + 2k}{6}\) = 5
2p + 3k = 30
- 2p - k = 6
\(\overline{\frac{4k}{4} = \frac{24}{4}}\)
k = 6
From 2p - k = 6
2p - 6 = 6
\(\frac{2p}{2} = \frac{12}{2}\)
p = 6
\(\frac{P}{2} + \frac{k}{3}\) = 5
\(\frac{3p + 2k}{6}\) = 5
2p + 3k = 30
- 2p - k = 6
\(\overline{\frac{4k}{4} = \frac{24}{4}}\)
k = 6
From 2p - k = 6
2p - 6 = 6
\(\frac{2p}{2} = \frac{12}{2}\)
p = 6