Question

The sum of five consecutive even integers is 1310. Find the integers

Answer 1

5 consecutive  even integers: n, n+2,n+4,n+6,n+8

n+(n+2)+(n+4)+(n+6)+(n+8)=1310

combine like terms

5n+20 = 1310

subtract 20 from each side

5n = 1290

divide by 5

n=258

n+2 = 260

n+4=262

n+6=264

n+8=266

Answer: 258,260,262,264,266

Answer 2

Answer:

Step-by-step explanation:

Let the first of these even numbers = n

The second one is n + 2

The third one is n + 4

The fourth one is n + 6

The fifth one is n + 8

Their sum is 1310

Equation

n + n + 2 + n + 4 + n + 6 + n + 8 = 1310

Solution

There are 5 ns.

The total of the numbers is 2 + 4 + 6 + 8 = 20

Simplify the given equation

5n +20 = 1310                        Subtract  20 from both sides.

5n + 20 - 20 = 1310 - 20       Combine

5n = 1290                               Divide by 5

5n/5 = 1290/5                         Divide

Answers

n1 = 258

n2 = 260

n3 = 262

n4 = 264

n5 = 266

Check

The total is 258 + 260 + 262 + 264 + 266 = 1310

It checks.