Find the largest of three consecutive whole numbers such that twice the sum of the two smallest numbers is 12 more than three times the larger number

Answer 1

For these types of problems, you need to break the parts down. Think about 3 consecutive whole numbers...for example, 12, 13, 14. Next, the words, “such that” means something that must be true. Next, “twice the sum” means 2 times the amount when you add.

twice the sum of the 2 smallest numbers have to be 12 more than 3 times the largest number.

Let’s used letters to represent the unknown numbers. If n is the first number, then the second number would be (n+1), and the third number would be (n+2). So twice the sum of the first two numbers would be

2[n + (n+1)] which equals 2(2n +1) = 4n+2

and 3 times the largest number would be 3(n+2) which equals

3n+6

Go back to the problem, twice the sum of the 2 smallest numbers is (“is” always means equal to)12 more than 3 times the larger number

Let’s put this into an algebraic equation...

4n+2 = 3n+6+12 now solve

4n +2=3n+18

n=16

therefore, the 3 consecutive whole numbers must be 16,17,18. Let’s double check.

twice the sum of the 2 smallest numbers is 2(16+17)=66

12 more than 3 times the largest number is 12+ 3(18)=66

Done!

Answer 2

x, x+1, x+2

2(x+x+1)=3(x+2)+12

2x+2x+2=3x+6+12

4x+2 =3x+18

x = 16

Answer 3

They are X,X+1,X+2

2(x + x+1) = 3(x+2)+12

2(2x+1) = 3(x+2)+12

4x+2 = 3x + 6+12

4x+2 = 3x+18

x+2=18

x=16

16,17,18

2(16+17) = 3(18)+12

2(33) = 54+12

66=66

YES

16,17,18