Let's solve the given equation:

\[
3475 \times x = x \times 145
\]

To isolate \( x \), first, we divide both sides by \( x \) (assuming \( x \neq 0 \)):

\[
3475 = 145
\]

This isn't possible unless \( x = 0 \). Hence, the only solution for this equation is:

\[
x = 0
\]

Would you like me to elaborate further on this, or do you have any questions?

Here are 5 related questions:
1. What would happen if \( x \neq 0 \) in this type of equation?
2. Can you solve for \( x \) in \( a \times x = b \times x \) for different values of \( a \) and \( b \)?
3. How can you check if an equation has multiple solutions?
4. What are the implications of having no valid solutions in an equation?
5. How do multiplication properties change when variables are introduced?

**Tip:** Always verify your assumptions, such as \( x \neq 0 \), before simplifying an equation!

Explore how to solve the linear equation 3475 × x = x × 145, discovering that the solution for x is 0. This problem involves understanding algebraic manipulation, the equality property of multiplication, and the importance of checking assumptions like x ≠ 0. Ideal for students in Grades 6-8.

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