document.write( "Question 144830: find the sum of 3 consecutive odd integers such that 4 times the sum of the 1st and second is 17 more than 7 times the 3rd \n" ); document.write( "
Algebra.Com's Answer #105517 by solver91311(24713)\"\" \"About 
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First odd integer: \"x\"
\n" ); document.write( "Second consecutive odd integer: \"x%2B2\"
\n" ); document.write( "Third consecutive odd integer: \"x%2B4\"\r
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\n" ); document.write( "\n" ); document.write( "The sum of the first and second: \"x+%2B+%28x+%2B+2%29=2x+%2B2\"
\n" ); document.write( "4 times the sum of the 1st and 2nd: \"4%282x%2B2%29=8x%2B8\"\r
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\n" ); document.write( "\n" ); document.write( "7 times the third: \"7%28x%2B4%29=7x%2B28\"
\n" ); document.write( "17 more than 7 times the 3rd: \"7x%2B28%2B17=7x%2B45\"\r
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\n" ); document.write( "\n" ); document.write( "So: \"8x%2B8=7x%2B45\". Solve this for x and verify that x is odd.\r
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\n" ); document.write( "\n" ); document.write( "The sum of the three integers is: \"x+%2B+%28x%2B2%29+%2B+%28x%2B4%29=3x%2B6\". Take the solution you found above, multiply by 3 and then add 6 to get your final answer.
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