Question

The sum of 3 consecutive integers is 18 more than 2 times the largest number. Find the three numbers. (Please help)

Answer 1

a+(a+1)+(a+2)=18+2(a+2)

a+a+1+a+2 = 18 + 2a + 4

3a + 3 = 22 + 2a

a + 3 = 22

a = 19

There, we found it.

Therefore, the integers are 19, 20, and 21.

Answer 2

19

20

21

Step-by-step explanation:

Givens

Let the first integer = x - 1

Let the second = x

Let the third = x + 1

Equation

x - 1 + x + x + 1 = 3x

Solution

3x = 2(x + 1) + 18 Remove the brackets

3x = 2x + 2 + 18 Combine the terms on the right

3x = 2x + 20 Subtract 2 x from both sides

x = 20

x - 1 = 19

x +1 = 21

Check

Sum = 60

2*Largest number = 42

18 more than 42 = 60