Given the following information: a is 20 more than 3 times the value of b . b is an even integer. ...which is true, regarding operations on the two numbers?A a + b and a × b will always be an odd integer.

Question

B a + b and a × b will always be an even integer.
C a + b will always be an odd integer, and a × b will always be an even integer.
D a + b will always be an even integer, and a × b will always be an odd integer.
E a + b and a × b will be rational numbers that are not in the set of integers.

Answer 1

To analyze the problem, we first express \( a \) in terms of \( b \):

Given: \[ a = 3b + 20 \]

Since \( b \) is an even integer, we can denote: \[ b = 2k \] for some integer \( k \).

Substituting into the equation for \( a \): \[ a = 3(2k) + 20 = 6k + 20 \] Notice that \( a \) is expressed as: \[ a = 6k + 20 = 2(3k + 10) \] This shows that \( a \) is also an even integer since it is divisible by 2.

Now we will analyze the operations \( a + b \) and \( a \times b \).

1. Calculate \( a + b \): \[ a + b = (6k + 20) + (2k) = 8k + 20 \] Because \( 8k + 20 \) is the sum of two even numbers (both \( 8k \) and \( 20 \) are even), \( a + b \) is even.

2. Calculate \( a \times b \): \[ a \times b = (6k + 20) \times (2k) = 12k^2 + 40k \] Here, both terms \( 12k^2 \) and \( 40k \) are clearly even, so \( a \times b \) is also even.

Conclusively, we find that:

• \( a + b \) is always even.
• \( a \times b \) is always even.

Thus, the correct answer is: B: \( a + b \) and \( a \times b \) will always be an even integer.