Question

I'm studying for my algebra exam and need help with a problem involving consecutive even integers where twice the first even integer is 20 more than the second even integer.

Answer 1

Let the first consecutive even integer be represented as \( x \). Since we are dealing with even integers, the next consecutive even integer can be represented as \( x + 2 \).

According to the problem, twice the first even integer is 20 more than the second even integer. We can set up the following equation based on that information:

\[
2x = (x + 2) + 20
\]

Now let's simplify the equation:

1. Distribute on the right side:
\[
2x = x + 2 + 20
\]

2. Combine like terms:
\[
2x = x + 22
\]

3. Now we can isolate \( x \) by subtracting \( x \) from both sides:
\[
2x - x = 22
\]
\[
x = 22
\]

So, the first even integer is \( 22 \).

Now, to find the second consecutive even integer:
\[
x + 2 = 22 + 2 = 24
\]

Thus, the two consecutive even integers are \( 22 \) and \( 24 \).

To verify the solution:
- Twice the first integer: \( 2 \times 22 = 44 \)
- Second integer plus 20: \( 24 + 20 = 44 \)

Since both calculations match, the solution is confirmed.

**Final answer: The consecutive even integers are 22 and 24.**