a) The numbers \(a, b, c\) are directly proportional to 3, 5, and 8 respectively. This means \(a = 3k\), \(b = 5k\), and \(c = 8k\) for some constant \(k\). Substitute these into the equation \(2a - 3b + 6c = 117\) and solve for \(k\). Then, find \(a, b, c\).
b) The numbers \(a, b, c\) are directly proportional to 5, 7, and 9 respectively. This means \(a = 5k\), \(b = 7k\), and \(c = 9k\) for some constant \(k\). Substitute these into the equation \(3a + 5b - 4c = 84\) and solve for \(k\). Then, find \(a, b, c\).
c) The numbers \(a, b, c\) are directly proportional to 5, 8, and 9 respectively. This means \(a = 5k\), \(b = 8k\), and \(c = 9k\) for some constant \(k\). Substitute these into the equation \(4a - 5b + 6c = 136\) and solve for \(k\). Then, find \(a, b, c\).
We need to determine the natural numbers \(a, b, c\) that are directly proportional to given sets of numbers and satisfy given linear equations. We will solve the first three sub-questions of the first problem.
For each sub-question, we will set up the proportional relationships:
- \(a = k_1 \cdot 3\)
- \(b = k_1 \cdot 5\)
- \(c = k_1 \cdot 8\)
where \(k_1\) is a constant of proportionality.
Given:
\[2a - 3b + 6c = 117\]
Substitute the proportional relationships:
\[2(3k_1) - 3(5k_1) + 6(8k_1) = 117\]
\[6k_1 - 15k_1 + 48k_1 = 117\]
\[39k_1 = 117\]
\[k_1 = \frac{117}{39} = 3\]
Using \(k_1 = 3\):
\[a = 3 \cdot 3 = 9\]
\[b = 3 \cdot 5 = 15\]
\[c = 3 \cdot 8 = 24\]
For the second sub-question:
- \(a = k_2 \cdot 5\)
- \(b = k_2 \cdot 7\)
- \(c = k_2 \cdot 9\)
Given:
\[3a + 5b - 4c = 84\]
Substitute the proportional relationships:
\[3(5k_2) + 5(7k_2) - 4(9k_2) = 84\]
\[15k_2 + 35k_2 - 36k_2 = 84\]
\[14k_2 = 84\]
\[k_2 = \frac{84}{14} = 6\]
Using \(k_2 = 6\):
\[a = 6 \cdot 5 = 30\]
\[b = 6 \cdot 7 = 42\]
\[c = 6 \cdot 9 = 54\]
For the third sub-question:
- \(a = k_3 \cdot 5\)
- \(b = k_3 \cdot 8\)
- \(c = k_3 \cdot 9\)
Given:
\[4a - 5b + 6c = 136\]
Substitute the proportional relationships:
\[4(5k_3) - 5(8k_3) + 6(9k_3) = 136\]
\[20k_3 - 40k_3 + 54k_3 = 136\]
\[34k_3 = 136\]
\[k_3 = \frac{136}{34} = 4\]
Using \(k_3 = 4\):
\[a = 4 \cdot 5 = 20\]
\[b = 4 \cdot 8 = 32\]
\[c = 4 \cdot 9 = 36\]
\[
\boxed{
\begin{aligned}
&\text{For the first equation:} \\
&a = 9, \quad b = 15, \quad c = 24 \\
&\text{For the second equation:} \\
&a = 30, \quad b = 42, \quad c = 54 \\
&\text{For the third equation:} \\
&a = 20, \quad b = 32, \quad c = 36 \\
\end{aligned}
}
\]