Question

The smaller of two consecutive even integers is five more than one half of the greater. Find the integers.

Answer 1

Answer:

12, 14

Step-by-step explanation:

Let's call the smallest number n.

Let n + 2 = the larger even integer.

Let's write our equation:

"The smaller of two consecutive even integers is five more" would be +5 and "nne half of the greater" can be written as $\frac{1}{2}$(n + 2).

n = 5 + $\frac{1}{2}$(n + 2)

Solvew for n.

n = 5 + $\frac{1}{2}$(n + 2)

Let's multiply each side by 2, to get rid of the fraction, $\frac{1}{2}$.

2 (n) = 2(5 + $\frac{1}{2}$(n + 2))

2n = 2* 5 + 2*($\frac{1}{2}$(n + 2))

2n = 10 + (n + 2)

2n = 10 + n + 2

2n = 10 + 2 + n

2n = 12 + n     Subtract n from each side.

2n - n = 12 + n - n

2n - n = 12

n = 12

Let's solve for our other integer:

n + 2 = 12 + 2 = 14

So our 2 consecutive, even integers are

12 and 14