https://file.aiquickdraw.com/mathbot/file/1730471668459gf5cti4v.jpg

The equation in the image is:

\[
\frac{x - 4}{6} + x = \frac{5x}{3} - \frac{x - 1}{18}
\]

To solve for \( x \), we can start by eliminating the fractions. The least common multiple of the denominators \(6\), \(3\), and \(18\) is \(18\). So, we can multiply every term by \(18\) to clear the fractions:

\[
18 \cdot \frac{x - 4}{6} + 18 \cdot x = 18 \cdot \frac{5x}{3} - 18 \cdot \frac{x - 1}{18}
\]

Simplify each term:

1. \(18 \cdot \frac{x - 4}{6} = 3(x - 4) = 3x - 12\)
2. \(18 \cdot x = 18x\)
3. \(18 \cdot \frac{5x}{3} = 6 \cdot 5x = 30x\)
4. \(18 \cdot \frac{x - 1}{18} = x - 1\)

This gives us the equation:

\[
3x - 12 + 18x = 30x - (x - 1)
\]

Simplify further:

\[
3x - 12 + 18x = 30x - x + 1
\]

Combine like terms:

\[
21x - 12 = 29x + 1
\]

Now, isolate \(x\) by moving terms involving \(x\) to one side and constants to the other side:

\[
-12 - 1 = 29x - 21x
\]

\[
-13 = 8x
\]

Divide by \(8\):

\[
x = -\frac{13}{8}
\]

### Final Answer:
\[
x = -\frac{13}{8}
\]

Would you like a more detailed breakdown of any step or have any questions?

### Related Questions
1. How would you solve if the equation had more complex fractions?
2. How can we verify the solution by substituting \( x = -\frac{13}{8} \) back into the original equation?
3. What if we had a variable in the denominator; how would that change the approach?
4. How do we find the least common multiple of several denominators efficiently?
5. Can this approach be applied to inequalities as well?

### Tip:
Always check for the least common multiple of denominators when clearing fractions in equations—it simplifies the process and minimizes calculation errors.

Solve the equation \( \frac{x - 4}{6} + x = \frac{5x}{3} - \frac{x - 1}{18} \).

Learn how to solve the equation \( \frac{x - 4}{6} + x = \frac{5x}{3} - \frac{x - 1}{18} \) by eliminating fractions and isolating \( x \). This step-by-step guide covers algebraic concepts such as fractions and the least common multiple, suitable for students in Grades 8-10.

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