document.write( "Question 526891: Find four consecutive odd integers such that 10 more than 3 times the third is 40 less than the first? \n" ); document.write( "

Algebra.Com's Answer #348714 by InfantileBear(2) $\"\"$ $\"About$

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To start off, write the numbers such that they are all in terms of the same number, such as:
\n" ); document.write( "x,x+2,x+4,and x+6;since consecutive odd integers are two numbers apart.
\n" ); document.write( "Now, if you work backwards, you subtract 40 from the first number:
\n" ); document.write( "x-40
\n" ); document.write( "and it also says that three times the third plus 10 is the first minus 40, so;
\n" ); document.write( "3(x+4)+10=3x+22
\n" ); document.write( "so altogether
\n" ); document.write( "3x-22=x+40
\n" ); document.write( "if you solve, you will get x, which x=31
\n" ); document.write( "so the three consecutive odd integers are 31, 31+2, 31+4, and 31+6
\n" ); document.write( "or 31,33,35,37 \n" ); document.write( "

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