Find 3 consecutive positive even integers such that the product of the second and third integer is twenty more than ten times the first integer. (Only algebraic solution, please help!)
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let x be the first number
let x + 2 be the second number
let x + 4 be the third number
(x + 2)(x + 4) = 10x + 20
factor out 10 from the right side.
(x + 2)(x + 4) = 10(x + 2)
you can cancel the factor (x + 2), which leaves
x + 4 = 10
x = 6
Therefore, the numbers are 6,8, and 10.