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| Question 1200102:  Find three consecutive even integers such that the product of the first and second is 20 less than 10 times the third
 Found 2 solutions by  math_tutor2020, josgarithmetic:
 Answer by math_tutor2020(3817)
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You can put this solution on YOUR website! Consecutive integers follow one after another.
 Examples:
 6,7,8
 22,23,24
 
 Consecutive even integers are when all of the values are even, and they are as close as possible.
 Examples:
 2,4,6
 100,102,104
 
 Let x be some integer. It might be even or odd.
 To guarantee it is always even, we double it to 2x.
 Eg:
 x = 5 doubles to 2x = 2*5 = 10
 x = 6 doubles to 2x = 2*6 = 12
 
 2x = first even integer
 2x+2 = even integer right after 2x
 2x+2+2 = 2x+4 = even integer right after 2x+2
 
 In short,
 first = 2x
 second = 2x+2
 third = 2x+4
 
 The sequence {2x,2x+2,2x+4} represents a set of three consecutive even integers, where x is any integer.
 
 We then have the following information " the product of the first and second is 20 less than 10 times the third".
 
 Let's translate that into an equation and solve for x.
 
 first*second = 10*third - 20
 2x*(2x+2) = 10*(2x+4) - 20
 4x^2+4x = 20x+40 - 20
 4x^2+4x = 20x+20
 4x^2+4x-20x-20 = 0
 4x^2-16x-20 = 0
 4(x^2-4x-5) = 0
 x^2-4x-5 = 0
 (x-5)(x+1) = 0
 x-5 = 0 or x+1 = 0
 x = 5 or x = -1
 
 Optionally you could use the quadratic formula to solve for x.
 Let me know if you need to see these steps.
 
 If x = 5, then,
 We end up with the set {10,12,14} as one solution.2x = 2*5 = 10 is the first even integer2x+2 = 2*5+2 = 10+2 = 12 is the second even integer2x+4 = 2*5+4 = 10+4 = 14 is the third even integer
 
 If x = -1, then,
 We end up with the set {-2,0,2} as the other solution.2x = 2*(-1) = -2 is the first even integer2x+2 = 2*(-1)+2 = -2+2 = 0 is the second even integer2x+4 = 2*(-1)+4 = -2+4 = 2 is the third even integer
 
 
 Check for the first solution:
 first*second = 10*third - 20
 10*12 = 10*14 - 20
 120 = 140 - 20
 120 = 120
 The first solution set is confirmed.
 
 Check for the second solution:
 first*second = 10*third - 20
 -2*0 = 10*2 - 20
 0 = 20 - 20
 0 = 0
 The second solution set is confirmed.
 
Answer by josgarithmetic(39630)
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