SOLUTION: find three consecutive even integers such that two times the sum of the first and the third integer is twenty-two less than the product of three and the second number

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Question 239617: find three consecutive even integers such that two times the sum of the first and the third integer is twenty-two less than the product of three and the second number
Found 2 solutions by College Student, checkley77:
Answer by College Student(505) About Me  (Show Source):
You can put this solution on YOUR website!
To find three concecutive even integers, we can say:
x = integer
x+2 = next consecutive even integer
x+4 = next to previous consecutive even integer
.
So, the equation would then be:
2%28x%2B%28x%2B4%29%29=%28x%2B2%29%28x%2B4%29-22
Now workout the algebra to solve for x. Once you do that, you'll now which two even consecutive numbers are the other two.
.
Remember to check your answer by substituting the x value in the original equation. Both sides of the equation must equal to each other.

Answer by checkley77(12569) About Me  (Show Source):
You can put this solution on YOUR website!
LET X, X+1 & X+2 BE THE THREE INTEGERS.
2(X+X+2)+22=3(X+1)
2(2X+2)+22=3X+3
4X+4+22=3X+3
4X-3X=3-22-4
X=-23 ANS. FOR THE FIRST INTEGER.
-23+1=-22 ANS. FOR THE MIDDLE INTEGER.
-23+2=-21 ANS. FOR THE LARGER INTEGER.
PROOF:
2(-23-21)+22=3*-22
2*-44+22=-66
-88+22=-66
-66=-66