https://file.aiquickdraw.com/mathbot/file/1727921101186is554oe5.jpg

Let's analyze the image step by step:

The image contains a set of three fractional equations written inside an oval shape:

1. \( \frac{1}{6} + \frac{2}{3} - \frac{1}{2} = 10 \)
2. \( \frac{3}{7} - \frac{3}{8} + \frac{2}{3} = 7 \)
3. \( \frac{2}{7} - \frac{1}{5} - \frac{3}{2} = 7 \)

Let's now solve each equation.

---

### Equation 1: \( \frac{1}{6} + \frac{2}{3} - \frac{1}{2} = 10 \)

To solve this, we first need to find a common denominator for the fractions \( \frac{1}{6} \), \( \frac{2}{3} \), and \( \frac{1}{2} \). The least common denominator (LCD) of 6, 3, and 2 is 6.

\[
\frac{1}{6} + \frac{2}{3} - \frac{1}{2} = 10
\]

Convert the fractions to have a denominator of 6:

\[
\frac{1}{6} + \frac{2}{3} \cdot \frac{2}{2} - \frac{1}{2} \cdot \frac{3}{3} = \frac{1}{6} + \frac{4}{6} - \frac{3}{6}
\]

Combine the fractions:

\[
\frac{1 + 4 - 3}{6} = \frac{2}{6} = \frac{1}{3}
\]

So the equation becomes:

\[
\frac{1}{3} = 10
\]

This is impossible because \( \frac{1}{3} \neq 10 \), so the first equation does not hold. There may be a mistake in the setup.

---

### Equation 2: \( \frac{3}{7} - \frac{3}{8} + \frac{2}{3} = 7 \)

To solve this equation, we need the least common denominator of 7, 8, and 3. The least common denominator is 168.

\[
\frac{3}{7} - \frac{3}{8} + \frac{2}{3} = 7
\]

Convert all fractions to have a denominator of 168:

\[
\frac{3}{7} = \frac{3 \times 24}{7 \times 24} = \frac{72}{168}
\]
\[
\frac{3}{8} = \frac{3 \times 21}{8 \times 21} = \frac{63}{168}
\]
\[
\frac{2}{3} = \frac{2 \times 56}{3 \times 56} = \frac{112}{168}
\]

Now substitute these into the equation:

\[
\frac{72}{168} - \frac{63}{168} + \frac{112}{168} = 7
\]

Simplify the left side:

\[
\frac{72 - 63 + 112}{168} = \frac{121}{168}
\]

So the equation becomes:

\[
\frac{121}{168} = 7
\]

This is also impossible since \( \frac{121}{168} \neq 7 \).

---

### Equation 3: \( \frac{2}{7} - \frac{1}{5} - \frac{3}{2} = 7 \)

The least common denominator of 7, 5, and 2 is 70.

Convert each fraction to have a denominator of 70:

\[
\frac{2}{7} = \frac{2 \times 10}{7 \times 10} = \frac{20}{70}
\]
\[
\frac{1}{5} = \frac{1 \times 14}{5 \times 14} = \frac{14}{70}
\]
\[
\frac{3}{2} = \frac{3 \times 35}{2 \times 35} = \frac{105}{70}
\]

Now substitute these into the equation:

\[
\frac{20}{70} - \frac{14}{70} - \frac{105}{70} = 7
\]

Simplify the left side:

\[
\frac{20 - 14 - 105}{70} = \frac{-99}{70}
\]

So the equation becomes:

\[
\frac{-99}{70} = 7
\]

Again, this is impossible since \( \frac{-99}{70} \neq 7 \).

---

### Conclusion

Each of the equations in the image seems to lead to contradictions or impossible results, indicating either errors in the setup or transcription of the equations. Let me know if you'd like further clarification or help with correcting these equations.

---

Do you have any questions, or would you like details on a specific part of the solution?

Here are 5 related questions to explore:

1. How do you find the least common denominator for a set of fractions?
2. What strategies can you use to simplify fractional equations?
3. How can you verify if a fractional equation is solvable?
4. What are the common errors in solving fractional equations?
5. How does multiplying both sides of an equation by the denominator help in clearing fractions?

**Tip**: Always check the initial setup of equations for any potential transcription errors before attempting to solve them.

Solve the system of fractional equations as presented in the image.

In this problem, we solve a set of fractional equations, focusing on concepts such as finding the least common denominator, combining fractions, and solving algebraic equations. This step-by-step guide helps understand how to handle complex fractions and solve systems of equations. Suitable for Grades 7-9.

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