Let's solve the equation \( 8x - 13 = 4r + 15 \).

Since this equation has both \( x \) and \( r \), I'll assume you're looking to isolate one variable (perhaps solving for \( x \) in terms of \( r \)).

### Steps:
1. Start by isolating the term with \( x \) on one side:
   \[
   8x = 4r + 15 + 13
   \]
2. Simplify the constants on the right-hand side:
   \[
   8x = 4r + 28
   \]
3. Solve for \( x \) by dividing both sides by 8:
   \[
   x = \frac{4r + 28}{8}
   \]
4. Simplify the expression:
   \[
   x = \frac{4(r + 7)}{8} = \frac{r + 7}{2}
   \]

So, the solution is:
\[
x = \frac{r + 7}{2}
\]

Would you like more details on the steps or have any further questions?

Here are 5 related questions to explore:
1. What happens if we solve for \( r \) in terms of \( x \)?
2. How can we check if the solution is correct by substituting values for \( r \)?
3. What if we wanted to solve the equation graphically?
4. How would the solution change if \( r \) were a known constant?
5. Can you express \( x \) in terms of other variables if additional terms were added?

**Tip:** Always double-check your final expression by substituting the variable values back into the original equation to verify the solution.

Learn how to solve the linear equation 8x - 13 = 4r + 15 with a step-by-step solution. This algebraic problem covers the process of isolating variables, simplifying expressions, and solving for one variable in terms of another. Suitable for students in Grades 8-10.

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