document.write( "Question 1108950: find three consecutive even integers such that the product of the first and third is 20 less than 10 times the second \n" ); document.write( "
Algebra.Com's Answer #723965 by addingup(3677)\"\" \"About 
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n-2 * n+2 = 10(n)-20
\n" ); document.write( "n-2 * n+2 = 10n - 20
\n" ); document.write( "n(n+2) + (-2)(n+2) = 10n - 20
\n" ); document.write( "n^2 + 2n + (-2)(n + 2) = 10n - 20
\n" ); document.write( "n^2 + 2n + ((-2)n +(-2) * 2) = 10n - 20
\n" ); document.write( "n^2 + 2n + (-2)n - 4 = 10n - 20
\n" ); document.write( "n^2 - 4 = 10n - 20
\n" ); document.write( "n^2 - 4 - (10n - 20) = 0
\n" ); document.write( "n^2 - 4 + ((-1) * 10n + (-20)(-1)) = 0
\n" ); document.write( "n^2 - 4 + ((-10)n + (-20)(-1) = 0
\n" ); document.write( "n^2 - 4 + (-10)n + 20 = 0
\n" ); document.write( "n^2 + 16 + (-10)n = 0
\n" ); document.write( "n^2 + (-10)n + 16 = 0
\n" ); document.write( "n^2 + (-10)n = -16
\n" ); document.write( "The coefficient of n is -10. Divide by 2 and square (-10/2 = -5; -5^2 = 25), then add result to both sides of the equation:
\n" ); document.write( "n^2 + (-10)n + 25 = 9
\n" ); document.write( "now we have the equation in the form x^2 + bx + c, a perfect square we can factor:
\n" ); document.write( "(n-5)^2 = 9
\n" ); document.write( "Take the square root of both sides:
\n" ); document.write( "n-5 = 3
\n" ); document.write( "n = 8
\n" ); document.write( "your numbers are:
\n" ); document.write( "n - 2, n, n + 2
\n" ); document.write( "= 8 - 2, 8, 8 + 2
\n" ); document.write( "= 6, 8, 10 Three even consecutive integers
\n" ); document.write( "the product of the first and third is 20 less than 10 times the second:
\n" ); document.write( "6(10) = 10(8) - 20
\n" ); document.write( "60 = 80 - 20 Correct\r
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