Solve the following addition problem by filling in the missing digits: _ 2 + 4 _ ----- 8 0

This is an addition problem where we need to find the missing digits. Let's represent the unknown digits with variables:

Let the first number be $$10x + 2$$, where $$x$$ is the digit in the tens place.

Let the second number be $$40 + y$$, where $$y$$ is the digit in the units place.

The sum is $$80$$.

So, the equation is:

\[ (10x + 2) + (40 + y) = 80 \]

Combining the known terms:

\[ 10x + y + 42 = 80 \]

Subtracting 42 from both sides:

\[ 10x + y = 80 - 42 \]

\[ 10x + y = 38 \]

From this equation, we can see that $$x$$ represents the tens digit and $$y$$ represents the units digit of the sum 38.

Therefore, $$x = 3$$ and $$y = 8$$.

Let's verify this by filling in the digits:

The first number is $$32$$.

The second number is $$48$$.

Adding them:

\[ 32 + 48 = 80 \]

The sum is correct.

Let's visualize this using column addition:

  3 2
+ 4 8
----- 
  8 0

In the units column, $$2 + 8 = 10$$. We write down $$0$$ and carry over $$1$$ to the tens column.

In the tens column, $$3 + 4 + 1$$ (carry-over) $$= 8$$. We write down $$8$$.

The result is $$80$$.

Ответ: 32 + 48 = 80