Let's solve the expression step by step:

Given expression:
\[
-10(5r + 8) + 3(3r + 4)
\]

### Step 1: Distribute the constants
- Distribute \(-10\) to both terms inside the parentheses:
  \[
  -10(5r) = -50r, \quad -10(8) = -80
  \]
  So, \(-10(5r + 8) = -50r - 80\).

- Distribute \(3\) to both terms inside the parentheses:
  \[
  3(3r) = 9r, \quad 3(4) = 12
  \]
  So, \(3(3r + 4) = 9r + 12\).

### Step 2: Combine the results
Now substitute the distributed values back into the expression:
\[
-50r - 80 + 9r + 12
\]

### Step 3: Combine like terms
- Combine the \(r\)-terms:
  \[
  -50r + 9r = -41r
  \]
- Combine the constant terms:
  \[
  -80 + 12 = -68
  \]

### Final simplified expression:
\[
-41r - 68
\]

Would you like more details or have any further questions?

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Here are five related questions to expand your understanding:
1. How would the result change if the constant term in the first parentheses was different?
2. Can you solve a similar expression like \(-6(2x + 3) + 4(5x - 1)\)?
3. What is the next step if you were asked to solve for \(r\) in the equation \(-41r - 68 = 0\)?
4. How would you handle distributing constants when one of the terms includes a square term like \(r^2\)?
5. What happens if the parentheses include fractions, like \(-3\left(\frac{5r}{2} + 8\right)\)?

**Tip**: When distributing, always remember to apply the constant to **both** terms inside the parentheses.

Learn how to simplify the algebraic expression -10(5r + 8) + 3(3r + 4) using the distributive property and combining like terms. This step-by-step guide covers key concepts in algebra for students in Grades 6-8.

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