SOLUTION: find three consecutive odd integers such that twice the second integer increased by six equals the sum of the largest and three time the smallest integer.

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Question 110493: find three consecutive odd integers such that twice the second integer increased by six equals the sum of the largest and three time the smallest integer.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Let us name these three unidentified integers x, y, and z, arbitrarily assuming that x is the smallest and z is the+largest, we may write:
if twice_+the_+second integer increased by six+equals_+the_+sum+of the largest_+and_+three_+time_+the_+smallest integer, we can write:

2y+%2B+6=+z+%2B+3x
If x is smallest odd number, then
y+=+x%2B2
z+=x%2B4
Now we can substitute it here:
2y+%2B+6=+z+%2B+3x
2+%28x%2B2%29+%2B+6=+x%2B4+%2B+3x
2+x+%2B+4+%2B+6=+x+%2B+4+%2B+3x

2+x+%2B+10+=+4x+%2B+4+….move 2x to the right and 4 to the left
+10++-+4+=+4x+%96+2x+….
+6++=++2x+….
+3++=++x+….

or +x++=++3+…. smallest odd number
y+=+x%2B2+=+3+%2B+2+=+5
z+=x%2B4+=+3+%2B+4+=+7
Check:
2y+%2B+6=+z+%2B+3x
2%2A5%2B+6=+7+%2B+3%2A3
10%2B+6=+7+%2B+9
16=+16