Show th above work on the sheet of paper

Let $S$ be the son's current age and $F$ be the father's current age.
Given:

$$S = 12 \text{ years}$$ $$F = S + 40$$

Substitute the son's age into the father's age equation:

$$F = 12 + 40$$ $$F = 52 \text{ years}$$

So, currently, the son is 12 years old and the father is 52 years old.

Let $x$ be the number of years from now when the father will be twice the son's age.
In $x$ years:
Son's age will be $S_x = S + x = 12 + x$.
Father's age will be $F_x = F + x = 52 + x$.
The condition is that the father's age will be twice the son's age:

$$F_x = 2 \times S_x$$ $$52 + x = 2(12 + x)$$

$$52 + x = 24 + 2x$$

Subtract $x$ from both sides of the equation:

$$52 = 24 + 2x - x$$ $$52 = 24 + x$$

Subtract 24 from both sides of the equation:

$$x = 52 - 24$$ $$x = 28$$

The father will be twice the son's age in $\boxed{28 \text{ years}}$.

We need to verify if the condition holds true in 28 years.
In 28 years:
Son's age will be $12 + 28 = 40$ years.
Father's age will be $52 + 28 = 80$ years.
Check if the father's age is twice the son's age:

$$80 = 2 \times 40$$ $$80 = 80$$

Since the equation holds true, the answer is proven correct.

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