To solve the system of equations by substitution, we have the following equations:
- \(\frac{3}{8}x + \frac{1}{3}y = \frac{17}{24}\)
- \(x + 7y = 8\)
First, let's solve the second equation for \(x\):
\[ x = 8 - 7y \]
Next, we can substitute this expression for \(x\) into the first equation:
\[ \frac{3}{8}(8 - 7y) + \frac{1}{3}y = \frac{17}{24} \]
Now, distribute \(\frac{3}{8}\):
\[ \frac{3 \cdot 8}{8} - \frac{3 \cdot 7y}{8} + \frac{1}{3}y = \frac{17}{24} \]
This simplifies to:
\[ 3 - \frac{21}{8}y + \frac{1}{3}y = \frac{17}{24} \]
To combine the \(y\) terms, we will need a common denominator for \(\frac{21}{8}y\) and \(\frac{1}{3}y\). The least common multiple of 8 and 3 is 24.
Convert \(\frac{21}{8}y\) to a denominator of 24:
\[ \frac{21}{8}y = \frac{21 \cdot 3}{8 \cdot 3}y = \frac{63}{24}y \]
Convert \(\frac{1}{3}y\) to a denominator of 24:
\[ \frac{1}{3}y = \frac{1 \cdot 8}{3 \cdot 8}y = \frac{8}{24}y \]
Now, substitute these back into the equation:
\[ 3 - \frac{63}{24}y + \frac{8}{24}y = \frac{17}{24} \]
Combine the \(y\) terms:
\[ 3 - \frac{63 - 8}{24}y = \frac{17}{24} \]
This simplifies to:
\[ 3 - \frac{55}{24}y = \frac{17}{24} \]
Next, isolate the \(y\) term. First, subtract 3 from both sides. Since \(3\) is the same as \(\frac{72}{24}\):
\[ -\frac{55}{24}y = \frac{17}{24} - \frac{72}{24} \]
This gives:
\[ -\frac{55}{24}y = \frac{17 - 72}{24} = -\frac{55}{24} \]
Now, we can multiply both sides by \(-\frac{24}{55}\) to solve for \(y\):
\[ y = 1 \]
Now that we have \(y\), substitute \(y = 1\) back into the equation for \(x\):
\[ x = 8 - 7(1) = 8 - 7 = 1 \]
So, we find:
\[ (x, y) = (1, 1) \]
Thus, the solution to the system of equations is:
\[ \boxed{(1, 1)} \]