Eric is twice as old as his brother Lucas. If 4 is subtracted from Eric's age and 4 is added to Lucas's age, their ages will be equal. What are the boys' ages now?

16 and 8.

16 and 8

Yes, that's correct!

Let's use algebra to check the answer:

- Let's call Eric's age "E" and Lucas's age "L".
- We're given that E = 2L, since Eric is twice as old as Lucas.
- We're also given that if we subtract 4 from Eric's age and add 4 to Lucas's age, they'll be equal: (E-4) = (L+4).
- We can substitute the first equation into the second equation: (2L-4) = (L+4).
- Simplifying that equation, we get L = 8.
- We can then use the first equation to find E: E = 2L = 2(8) = 16.

So we get the same answer of 16 and 8!

Let's break down the information given:

1. Eric is twice as old as Lucas: Let's assume Lucas's age is represented by x. So, Eric's age is 2x.

2. If 4 is subtracted from Eric's age and 4 is added to Lucas's age, their ages will be equal: This means that (2x - 4) = (x + 4).

Now, let's solve the equation to find the value of x, which is Lucas's age:

2x - 4 = x + 4

First, we can subtract x from both sides to isolate the x term:

2x - x - 4 = x - x + 4

This simplifies to:

x - 4 = 4

Next, we can move the constant term to the other side of the equation by adding 4 to both sides:

x - 4 + 4 = 4 + 4

This simplifies to:

x = 8

Therefore, Lucas's age (x) is 8 years old. Since Eric is twice as old, his age is 2x, which is 2 * 8 = 16 years old.

So, Eric is 16 years old, and Lucas is 8 years old.

Ms Sue please help me!