David is 6 years older than his friend Sam. Five years ago, David was twice as old as Sam. How old are they now?

Let's define the variables for the ages of David and Sam:

- Let \( D \) be David's current age.
- Let \( S \) be Sam's current age.

We are given two pieces of information:

1. David is 6 years older than Sam:
   \[
   D = S + 6
   \]

2. Five years ago, David was twice as old as Sam:
   \[
   D - 5 = 2(S - 5)
   \]

### Solving the Equations

Substitute the first equation into the second:

\[
(S + 6) - 5 = 2(S - 5)
\]

Simplify the equation:

\[
S + 1 = 2S - 10
\]

Rearrange to solve for \( S \):

\[
1 + 10 = 2S - S \implies S = 11
\]

Now, substitute \( S = 11 \) back into the first equation to find \( D \):

\[
D = S + 6 = 11 + 6 = 17
\]

### Conclusion

- **Sam is 11 years old.**
- **David is 17 years old.**

### Verification

To verify, check the condition given in the problem:

- Five years ago, Sam was \( 11 - 5 = 6 \) years old.
- Five years ago, David was \( 17 - 5 = 12 \) years old.

Since 12 is indeed twice 6, the solution is correct.

Would you like more details or have any questions?

### Related Questions:

1. If David was 3 times older than Sam 10 years ago, what would be their current ages?
2. How old will David be when he is twice Sam's current age?
3. If Sam’s age 5 years from now is half of David's current age, what is David’s age now?
4. If David and Sam’s ages sum up to 40, what are their current ages?
5. How many years from now will David be twice as old as Sam?

**Tip:** Always double-check by substituting your values back into the original conditions.

Learn how to solve a classic age problem involving linear equations: David is 6 years older than Sam, and 5 years ago, David was twice as old as Sam. Discover the step-by-step solution to find their current ages. Suitable for students in Grades 6-8.

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