SOLUTION: Solve algebraically using one variable. Find three consecutive even integers such that the product of the smallest integer and the middle integer is eight less than eleven times th

Question 86104: Solve algebraically using one variable. Find three consecutive even integers such that the product of the smallest integer and the middle integer is eight less than eleven times the largest integer. Thanks for the help.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Since consecutive even integers are two numbers apart (think of 2, 4, and 6), then if you
let x be the first even integer, the next consecutive even integer will be x + 2 and the
third even integer in this consecutive series will be x + 4. In summary:
.
Smallest = x
Middle = x + 2
Largest = x + 4
.
The product of the smallest and the middle integers is represented by:
.
x*(x + 2)
.
Eleven times the largest integer is represented by:
.
11*(x + 4)
.
Since the product of the smallest and middle integers is 8 less than 11 times the largest
integer, if we add 8 to the product of the smallest and middle integers the result will
be equal to 11 times the largest. In equation form this equality becomes:
.
x*(x + 2) + 8 = 11*(x + 4)
.
That's the equation in one variable (x). Now it just needs to be solved. Begin by doing
the two distributed multiplications ... the one on the left side tells you to multiply
x times both quantities inside the parentheses and the one on the right side tells you
to multiply 11 times both quantities inside its parentheses. As a result of those two
multiplications the equation becomes:
.
x^2 + 2x + 8 = 11x + 44
.
The x^2 term tells you that the equation is a quadratic and the form to get a quadratic
equation into for solving is:
.
ax^2 + bx + c = 0
.
That means we need to remove everything from the right side of the equation leaving zero
on the right side and then combine like terms on the left side.
.
Begin by subtracting 11x from the right side. But to maintain the equality, when you do that
you must also subtract 11x from the left side. This subtraction on both sides results in
the 11x disappearing from the right side and appearing on the left side, and the equation becomes:
.
x^2 + 2x + 8 - 11x = + 44
.
Do the same process for the +44 on the right side. Subtract 44 from both sides and the
resulting equation is:
.
x^2 + 2x + 8 - 11x - 44 = 0
.
Getting close to what is needed. Now combine the +2x and -11x which results in -9x and
also combine the +8 and the -44 which results in -36. This simplifies the equation to:
.
x^2 - 9x - 36 = 0
.
We can solve this equation by factoring. Notice that factors of -36 are -12 and +3 and
when they are added together the result is -9. Therefore, the left side of this equation can
be factored to give:
.
(x - 12)*(x + 3) = 0
.
Next notice that the left side of this equation will equal zero if either of the factors are
zero. Therefore, the equation will work if either x- 12 = 0 or if x + 3 = 0. So solving
these two equations tells you two numbers that will work for x:
.
x - 12 = 0 is solved by adding 12 to both sides to get x = 12, and
.
x + 3 = 0 is solved by subtracting 3 from both sides to get x = -3.
.
But remember that the solution is to be EVEN integers. Therefore, -3 (which is odd) can be
ignored. The answer answer then becomes:
.
Smallest integer = x = 12
Middle integer = x + 2 = 12 + 2 = 14
Largest integer = x + 4 = 12 + 4 = 16
.
Check:
.
The product of the smallest and middle integer is 12 * 14 = 168
.
Add 8 to this product and the result is 176
.
11 times the largest integer is 11 * 16 = 176
.
Everything works OK.
.
In summary the 3 consecutive even integers that satisfy the problem are 12, 14, and 16
.
Hope this helps you to understand the problem and one way to solve it (by factoring).
You could also have used the quadratic formula to solve it, but factoring was probably
the easiest once you recognize that the left side can be factored.

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