Find three consecutive integers

Question

the sum of twice the smallest integer and 4 times the largest integer is equal to 8 more than 5 times the middle integer.

Answer 1

Think of them as x, y , and z. 

x+1=y

y+1=z 

x+2=z

These are all properties if they are consecutive integers. 

Then you know that the sum of 2x and 4z = 5y+8

so 2x+4z=5y+8. 

Then plug in the equations for y and z in terms of x: 

2x+4(x+2)=5(x+1)+8

Solve for x and then you can find the other two integers too. 

Answer 2

Let the smallest integer be x.

Then because the numbers are consecutive:

The middle integer is x+1.

The highest integer is x+2. 

 

From the problem description, the following  is known:

twice the smallest integer is 2x

4 times the largest integer is (4)(x+2)

8 more than 5 times the middle integer is (5)(x+1) + 8

 

Based on the problem description, the following equation is made:

2x + (4)(x+2) = (5)(x+1) + 8

 

Solving this equation for x:

2x + 4x + 8 = 5x + 5 + 8

6x + 8 = 5x + 13

x = 13 - 8 = 5

 

Therefore, the smallest integer would be 5 and the three consecutive numbers are: 5, 6, and 7.