Solve algebraically using one variable: Find three consecutive odd integers such that the

Question

product of the first integer and the third integer is equal to nine more than twelve times the middle integer.

Answer 1

Every odd integer can be written in the form 2 k + 1

So your first integer is a1 = 2 k + 1

Second integer is a2 = first integer + 2 = a1 + 2 = 2 k + 1 + 2 = 2 k + 3

Third integer is a3 = second integer + 2 = a2 + 2 = 2 k + 3 + 2 = 2 k + 5

Product of the first integer and the third integer is equal to nine more than twelve times the middle integer mean:

a1 * a3 = 12 * a2 + 9

Replace a1 = 2 k + 1 , a2 = 2 k + 3 and a3 = 2 k + 5 in this equation.

( 2 k + 1 ) * ( 2 k + 5 ) = 12 * ( 2 k + 3 ) + 9

2 k * 2 k + 2 k * 1 + 5 * 2 k + 5 * 1 = 12 * 2 k + 12 * 3 + 9

4 k ^ 2 + 2 k + 10 k + 5 = 24 k + 36 + 9

4 k ^ 2 + 12 k + 5 = 24 k + 45 Subtract 24 k to both sides

4 k ^ 2 + 12 k + 5 - 24 k = 24 k + 45 - 24 k

4 k ^ 2 - 12 k + 5 = 45 Subtract 45 to both sides

4 k ^ 2 - 12 k + 5 - 45 = 45 - 45

4 k ^ 2 - 12 k - 40 = 0 Divide both sides by 4

k ^ 2 - 3 k - 10 = 0

Try to solve this quadratic equation.

The solutions are:

k = - 2 and k = 5

You have two set of the solutions:

1)

k = - 2

a1 = 2 k + 1 = 2 * ( - 2 ) + 1 = - 4 + 1 = - 3

a2 = a1 + 2 = - 3 + 2 = - 1

a3 = a2 + 2 = - 1 + 2 = 1

- 3 , - 1 , 1

2)

k = 5

a1 = 2 k + 1 = 2 * 5 + 1 = 10 + 1 = 11

a2 = a1 + 2 = 11 + 2 = 13

a3 = a2 + 2 = 13 + 2 = 15

11 , 13 , 15