find three consecutive integers such that the sum of one-half of the smallest and one third of the largest is one less than the other integer.

Question

I was asked to form an equation and then solve from the description i was given but i am stuck after i form the equation.

Answer 1

Let x = smallest integer

x+1 = middle integer

x+2 = largest integer

 

From the statement:

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

 

Distribute 1/3 to x+2:

1/2x + 1/3x + 2/3 + 1 = x+1

 

1 appears on both sides, so we can cancel them out.

 

1/2x + 1/3x + 2/3 = x

 

Make a common denominator of 6:

3/6x + 2/6x + 4/6 = 6/6x

 

Combine like terms on the left:

5/6x + 4/6 = 6/6x

 

Move 5/6x to the right:

4/6 = 1/6x

 

Multiply by 6 to cancel out the fraction:

4 = x

5 = x+1

6 = x+2