Find four consecutive integers such that twice the sum of the first and third is 11 greater than 3 times the second.

Question

-four consecutive integers

-twice the sum of the first and third

-11 greater than 3 times the second

Answer 1

1st integer=x

2nd integer=x+1

3rd integer=x+2

4th integer=x+3

 

2(x +(x+2))= 3(x+1) +11

 

solve

 

2 (2x +2) = 3x + 3 + 11

 

4x + 4     = 3x + 14

 

-3x-4       =-3x -4

 

 

            x=10

 

1st integer=10

2nd integer=11

3rd integer=12

4th integer=13

 

 

Answer 2

Here's what we know:

we have a, b, c, d

2(a + c) = 3b + 11

b = a + 1

c = a + 2

d = a + 3

 

2(a + a + 2) = 3(a + 1) + 11

2(2a + 2) = 3a + 3 + 11

4a + 4 = 3a + 14

Subtract 3a from both sides:

a + 4 = 14

Subtract 2 from both sides:

a = 10

 

10, 11, 12, 13

 

Check:

2(10 + 12) = 11 + 3*11

2 * 22 = 11 + 33

44 = 44

Answer 3

The four integers n, n+1, n+2, and n+3

 

Twice the sum of the first and third is 2*(n+n+2)

11 greater than 3 times the second is: 11+3*(n+1)

 

Then: 2*(2*n+2) = 11+3*(n+1)

2*(2*n+2)=2*2*(n+1)=4*(n+1)

 

So: 4*(n+1)=11+3*(n+1)

      (n+1)=11

      n=10

Then the 4 integers are: 10, 11, 12, and 13

 

Check: Twice sum of 1st and 3rd is: 2*(10+12)=2*22=44

           11 greater than 3 times the second is: 11+3*11 = 11+33=44