https://file.aiquickdraw.com/mathbot/file/17258005791578mlc3xrp.jpg

The image shows an equation that needs to be proven under the condition \( a : b = c : d \). The equation is:

\[
\frac{2a - 5b}{2a + 5b} = \frac{2c - 5d}{2c + 5d}
\]

### Given:
\[
a : b = c : d \quad \text{or equivalently} \quad \frac{a}{b} = \frac{c}{d}
\]

We need to prove that:
\[
\frac{2a - 5b}{2a + 5b} = \frac{2c - 5d}{2c + 5d}
\]

### Proof:
1. Start with the proportion \( \frac{a}{b} = \frac{c}{d} \). From this, we can write:
   \[
   a = \lambda c \quad \text{and} \quad b = \lambda d \quad \text{for some constant } \lambda.
   \]
   
2. Substitute these values of \( a \) and \( b \) into the left-hand side of the equation:

\[
\frac{2a - 5b}{2a + 5b} = \frac{2(\lambda c) - 5(\lambda d)}{2(\lambda c) + 5(\lambda d)} = \frac{\lambda(2c - 5d)}{\lambda(2c + 5d)}
\]

3. Simplify the expression:
   \[
   \frac{\lambda(2c - 5d)}{\lambda(2c + 5d)} = \frac{2c - 5d}{2c + 5d}
   \]

Thus, the left-hand side simplifies to the right-hand side, proving that:

\[
\frac{2a - 5b}{2a + 5b} = \frac{2c - 5d}{2c + 5d}
\]

This completes the proof.

---

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:
1. How does the concept of ratios help in solving such equations?
2. Can you apply a similar method to other types of proportional equations?
3. What happens if the ratio \( a : b \neq c : d \)? How would the equation change?
4. Can you solve this equation without assuming \( a = \lambda c \) and \( b = \lambda d \)?
5. How could this method be applied to real-world problems involving proportions?

**Tip:** Always check if variables in proportional equations can be expressed using a common factor. It simplifies complex expressions easily.

If a : b = c : d, then prove that (2a - 5b) / (2a + 5b) = (2c - 5d) / (2c + 5d).

This page provides a detailed proof of how to solve an algebraic equation involving ratios and proportions: (2a - 5b) / (2a + 5b) = (2c - 5d) / (2c + 5d), given that a : b = c : d. Suitable for students in Grades 10-12.

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