Rational Equations

Question

There were 3 consecutive even integers. Write a variable for the first number. Write a phrase for half of the second number. Write a phrase for a third of the third number. All three phrases added together will equal 17. What are the numbers? [Hint: integers- x, x+2, x+4]

Answer 1

While Joseph got the correct answer the problem was not actually worked out in quite the correct manner.  Note that in order to be certain the 1st number was even as opposed to odd the first number in our sequence should actually be represented by 2x instead of x since 2 times any number will always be even.  This gives us the 3 numbers as being

2x, 2x + 2, and 2x + 4 for 3 consecutive even integers.  In actuality the hint given was not actually correctly stated in the original problem.

 

1/2 of the 2nd term is (2x+2)/2 = x+1

1/3 of the 3rd term is (2x+4)/3

 

Since the 1st term plus 1/2 2nd term plus 1/3 of 3rd term add to 17 we have

 

2x + x + 1 + (2x+4)/3 = 17

Multiply all by 3 to get rid of the fraction

6x + 3x + 3 + 2x + 4 = 51

11x + 7 = 51

11x = 44

x = 4

 

Note our first term is actually 2x so the 1st even integer is 2(4) = 8

 

Thus, the 3 consecutive even integers are 8, 10, 12

Answer 2

let x = the first even integer; then the second integer will be x + 2, and the third integer will be x + 4

 

We are asked to write a phrase for half of the second integer which will be (x + 2)/2

 

We are asked to write a phrase for the third integer which will be (x + 4)/3

 

we are told that their sum is 17, so x + (x+2)/2 + (x+4)/3 = 17

 

we multiply each element by 6 to get 6x + 3(x+2) + 2(x+4) = 17

 

we expand the parentheses to get 6x + 3x + 6 + 2x + 8 = 102

 

we combine the like terms to get 11x + 14 = 102

 

we solve for x, which equals 8

 

so, (x+2)/2 = 10/2 = 5; and (x + 4)/3 = 12/3 = 4

 

Proof: 8 + 5 + 4 = 17