Solve the system of equations by substitution.

x + y = A system of equations. StartFraction 3 over 8 EndFraction x plus StartFraction one-third EndFraction y equals StartFractions 17 over 24 EndFraction.

x + 7y = 8

To solve the given system of equations by substitution, let's first rewrite them in a clear format:

1. \( x + y = \frac{3}{8} \)
2. \( \frac{1}{3}x + \frac{1}{3}y = \frac{17}{24} \)
3. \( x + 7y = 8 \)

### Step 1: Solve the first equation for \( x \)
The first equation is:

\[ x + y = \frac{3}{8} \]

Solve for \( x \):

\[ x = \frac{3}{8} - y \]

### Step 2: Substitute \( x = \frac{3}{8} - y \) into the second equation

The second equation is:

\[ \frac{1}{3}x + \frac{1}{3}y = \frac{17}{24} \]

Substitute \( x = \frac{3}{8} - y \) into this equation:

\[
\frac{1}{3} \left( \frac{3}{8} - y \right) + \frac{1}{3} y = \frac{17}{24}
\]

Distribute \( \frac{1}{3} \) across the terms inside the parentheses:

\[
\frac{1}{3} \times \frac{3}{8} - \frac{1}{3} y + \frac{1}{3} y = \frac{17}{24}
\]

Simplify the terms:

\[
\frac{1}{8} + 0 = \frac{17}{24}
\]

Now we have:

\[
\frac{1}{8} = \frac{17}{24}
\]

But \( \frac{1}{8} \) is not equal to \( \frac{17}{24} \), so there is a contradiction.

### Conclusion
The system of equations does not have a solution. The two equations are inconsistent, and no values for \( x \) and \( y \) will satisfy both equations at the same time.

Would you like any details or have any questions about this?

### Relative Questions:
1. How do we identify when a system of equations has no solution?
2. Can we determine if a system has infinitely many solutions using substitution?
3. What are other methods to solve systems of equations besides substitution?
4. How can we verify the solution of a system of equations?
5. What is the difference between inconsistent and dependent systems?

### Tip:
When solving systems of equations, always double-check your substitutions to avoid errors that might lead to contradictions or incorrect results.

Explore the process of solving a system of equations by substitution. This example demonstrates an inconsistent system, where no solution exists. Suitable for students in Grades 8-10.

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