SOLUTION: Three brothers have ages that are consecutive even intergers. The product of the first and third boys' ages is 20 more than twice the second boy's age. Find the age of each of th

Question 66790: Three brothers have ages that are consecutive even intergers. The product of the first and third boys' ages is 20 more than twice the second boy's age. Find the age of each of the three boys.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=first brother's age
x+2=second brother's age
x+4=third brother's age
Now we are told that the product of the first and third boys' age x(x+4) is 20 more than twice the second boy's age(20+2(x+2)). So our equation to solve is:
x(x+4)=20+2(x+2)
x^2+4x=20+2x+4 subtract 2x and 24 from both sides
x^2+4x-2x-24=24-24+2x-2x collecting like terms:
x^2+2x-24=0 This can be factored:
(x+6)(x-4)=0
x=-6 discount this solution---age is not negative

x=4 ------------age of the first brother
x+2=4+2=6-----------------age of the second brother
x+4=4+4=8--------------------age of the third brother
CK
Product of 1st and 3rd boys age is 20 more than twice the product of the second boys age
Product of 1st and 3rd boys age =4*8=32
20 more than twice the product of the second boy's age is 20+12=32
4---6----8=consecutive even integers ck

Hope this helps----ptaylor

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