Question 161590: Find three consecutive odd integers such that the product of the first and second integer is 7 less then 10 times the third integer.
Found 2 solutions by nerdybill, checkley77:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Find three consecutive odd integers such that the product of the first and second integer is 7 less then 10 times the third integer.
.
Let x = first consecutive odd integer
then
x+2 = second consecutive odd integer
x+4 = third consecutive odd integer
.
From: "product of the first and second integer is 7 less then 10 times the third integer"
x(x+2) = 10(x+4) - 7
x^2 + 2x = 10x + 40 - 7
x^2 + 2x = 10x + 33
x^2 - 8x - 33 = 0
.
Factoring gives us:
(x-11)(x+3) = 0
therefore,
x = {-3, 11}
.
Two possible solutions:
11, 13, 15
-3, -1, 1
Answer by checkley77(12844)  (Show Source):
You can put this solution on YOUR website! Let x, x+2 & x+4 be the 3 integers.
x(x+2)=10(x+4)-7
x^2+2x=10x+40-7
x^2+2x-10x-40+7=0
x^2-8x-33=0
(x-11)(x+3)=0
x-11=0
x=11 answer
x+3=0
x=-3 answer.
The 3 integers are:11,13,15 or:
The 3 integers are: -3,-1,& +1.
Proofs:
11*13=10*15-7
143=150-7
143=143
-3*-1=10*1-7
3=10-7
3=3
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