The age difference between a sister and a sibling is 8 years. Four years later, the sum of their ages is equal to twice the sum of their ages 2 years ago. Find the sister's current age.

1. Problem Description

The age difference between a sister and a sibling is 8 years. Four years later, the sum of their ages is equal to twice the sum of their ages 2 years ago. Find the sister's current age.

2. Solution Steps

Let $x$ be the sister's current age, and $y$ be the sibling's current age.

From the first sentence, we have:

$x - y = 8$

So, $y = x - 8$.

Four years later, the sister's age is $x + 4$, and the sibling's age is $y + 4$. The sum of their ages is $(x + 4) + (y + 4) = x + y + 8$.

Two years ago, the sister's age was $x - 2$, and the sibling's age was $y - 2$. The sum of their ages was $(x - 2) + (y - 2) = x + y - 4$.

From the second sentence, we have:

$x + y + 8 = 2(x + y - 4)$

$x + y + 8 = 2x + 2y - 8$

$x + y = 16$

Substitute $y = x - 8$ into the equation $x + y = 16$:

$x + (x - 8) = 16$

$2x - 8 = 16$

$2x = 24$

$x = 12$

So, the sister's current age is

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2. The sibling's current age is $y = x - 8 = 12 - 8 = 4$.

Four years later, the sister will be $12 + 4 = 16$, and the sibling will be $4 + 4 = 8$. The sum of their ages will be $16 + 8 = 24$.

Two years ago, the sister was $12 - 2 = 10$, and the sibling was $4 - 2 = 2$. The sum of their ages was $10 + 2 = 12$.

$24 = 2 \times 12$

3. Final Answer

The sister's current age is

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2.