Question 225514
Find three consecutive even integers such that the sum of the first integer and three times the third integer is 64 more than the second integer.


Step 1.  Let n be the first even integer.


Step 2.  Let n+2 and n+4 be the next two consecutive even integers.


Step 3.  Let 3(n+4) be three times the third integer.


Step 4.  Let n+3(n+4) be the sum of the first integer and 3 times the third integer


Step 5.  Let n+2+64=n+66 be 64 more than the second integer.


Step 6.  Then, n+3(n+4)=n+66 since the sum of the first integer and three times the third integer is 64 more than the second integer


Step 7.  Solving yields the following steps


*[invoke explain_simplification "n+3(n+4)=n+66" ]


With n=18, then n+2=20 and n+4=22


Check equation in Step 6:  n+3(n+4)=n+66...18+3*22=18+66 or 84=84...which is a true statement.


Step 8  The three consecutive even integers are 18, 20, and 22.


I hope the above steps were helpful.


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And good luck in your studies!


Respectfully,
Dr J

http://www.FreedomUniversity.TV