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!)
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.
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