document.write( "Question 196765: what is the sum of three consecutive odd integers if the sum of the second and third numbers is six more than two times the first number? \n" ); document.write( "

Algebra.Com's Answer #147474 by Earlsdon(6294) $\"\"$ $\"About$

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Let (2x-1) be the first odd number, ( x can be any integer), so (2x-1+2 = 2x+1) will be the next cosecutive odd integer and (2x-1+4 = 2x+3) will be the third consecutive odd integer.
\n" ); document.write( "From the problem description, we have:
\n" ); document.write( "(2x+1)+(2x+3) = 2(2x-1)+6 Simplifying, we have:
\n" ); document.write( "4x+4 = 4x+4
\n" ); document.write( "Well, as you can see, we have an identity which means that x can be any value.
\n" ); document.write( "So the answer to the problem is that you can pick any three consecutive odd integers at random and their sum will be the solution.
\n" ); document.write( "For example:
\n" ); document.write( "17, 19, and 21
\n" ); document.write( "19+21 = 2(17)+6
\n" ); document.write( "40 = 34+6
\n" ); document.write( "40 = 40 \n" ); document.write( "

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