document.write( "Question 525: I always have problems with these word problems: Find three consecutive even numbers so that the sum of the first and third numbers is 22 less than three times the second number. This one is hard, I would love it if you helped me out on this please and thank you!\r
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Algebra.Com's Answer #197 by arden42(16)\"\" \"About 
You can put this solution on YOUR website!
Well, first we need to assign letters to each of the numbers. x, y and z will do.\r
\n" ); document.write( "\n" ); document.write( "Now, if they are consecutive even numbers, then we know that x is 2 less than y, and y is 2 less than z:\r
\n" ); document.write( "\n" ); document.write( "\"x+=+y-2\"
\n" ); document.write( "\"y+=+z-2\"\r
\n" ); document.write( "\n" ); document.write( "We'll rearrange the second one to be consistent:\r
\n" ); document.write( "\n" ); document.write( "\"z+=+y+%2B+2\"\r
\n" ); document.write( "\n" ); document.write( "Now for the rest of the question. The first number plus the third number is 22 less than the second number multiplied by 3:\r
\n" ); document.write( "\n" ); document.write( "\"x+%2B+z+=+3y+-+22\"\r
\n" ); document.write( "\n" ); document.write( "To work out the numbers, we will need to substitute the first 2 equations into the third, like this:\r
\n" ); document.write( "\n" ); document.write( "\"%28y+-+2%29+%2B+%28y+%2B+2%29+=+3y-22\"\r
\n" ); document.write( "\n" ); document.write( "Expanding the brackets gives:\r
\n" ); document.write( "\n" ); document.write( "\"2y+=+3y-22\"\r
\n" ); document.write( "\n" ); document.write( "Adding 22 to both sides, and taking away 2y from both sides:\r
\n" ); document.write( "\n" ); document.write( "\"22=y\"\r
\n" ); document.write( "\n" ); document.write( "So the middle number is 22. Since the other numbers are the even numbers either side of it, the final answer is:\r
\n" ); document.write( "\n" ); document.write( "The 3 numbers are 20, 22 and 24.
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