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?

1 answer

Let's break down the given information step by step.

We have:

  • \( a = 3b + 20 \)
  • \( b \) is an even integer.

Since \( b \) is an even integer, it can be expressed in the form \( b = 2k \) where \( k \) is an integer.

Substituting \( b = 2k \) into the equation for \( a \):

\[ a = 3(2k) + 20 = 6k + 20 \]

Now we have both \( a \) and \( b \) expressed in terms of \( k \):

  • \( a = 6k + 20 \)
  • \( b = 2k \)

Next, we can analyze their properties:

  1. Both \( a \) and \( b \) are integers: Since both \( k \) and the operations performed on \( k \) (multiplication and addition) yield integers, \( a \) and \( b \) are integers.

  2. Parity:

    • \( b \) is even by definition.
    • For \( a = 6k + 20 \):
      • Both \( 6k \) and \( 20 \) are even, so the sum \( 6k + 20 \) is also even.
    • Thus, both \( a \) and \( b \) are even integers.
  3. Relationship:

    • Since \( a = 3b + 20 \), \( a \) is always 20 units greater than three times \( b \). This means \( a \) will always be larger than \( b \) for any positive \( b \).

In conclusion, the following can be stated:

  • Both \( a \) and \( b \) are even integers.
  • \( a \) is always greater than \( b \) (unless \( b \) is sufficiently negative such that the equation results in a negative \( a \)).
  • The specific operations you can perform are addition, subtraction, and multiplication, with results that will always maintain the even nature.

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