Given that S is between R and T, RS = x + 3, ST = 8x, and RT = 57. Find the value of x.


We are given that point \( S \) is between points \( R \) and \( T \), and the following conditions:

- \( RS = x + 3 \)
- \( ST = 8x \)
- \( RT = 57 \)

Since \( S \) is between \( R \) and \( T \), the sum of the lengths of \( RS \) and \( ST \) equals \( RT \). Therefore, we can set up the equation:

\[
RS + ST = RT
\]

Substituting the given expressions for \( RS \), \( ST \), and \( RT \):

\[
(x + 3) + 8x = 57
\]

Now, combine like terms:

\[
9x + 3 = 57
\]

Subtract 3 from both sides:

\[
9x = 54
\]

Finally, divide by 9:

\[
x = 6
\]

Thus, the value of \( x \) is \( \boxed{6} \).

Would you like more details on this process? Here are some related questions for further practice:

1. If \( RS = 6x \) and \( ST = x + 4 \), and \( RT = 40 \), find \( x \).
2. If \( x = 6 \), calculate the lengths of \( RS \) and \( ST \).
3. What is the relationship between segment addition and point location?
4. How do you solve for a variable in a linear equation?
5. How would this problem change if \( ST \) were given as \( 10x \)?

**Tip:** When solving geometry problems involving segment addition, always check if you can set up an equation where the total length equals the sum of the parts.

Given that S is between R and T, RS = x + 3, ST = 8x, and RT = 57. Find the value of x.

Find the value of x in a linear equation problem where RS = x + 3, ST = 8x, and RT = 57. This example covers key algebraic and geometric concepts like segment addition and solving for variables, suitable for students in Grades 6-8.

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