document.write( "Question 21115: I've been trying to help my son with this problem but can't figure out where to start. The problem is this, Twice the greater of two consecutive odd integers is 13 less than three times the lesser number. Find the integers. \n" ); document.write( "

Algebra.Com's Answer #10148 by Earlsdon(6294) $\"\"$ $\"About$

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Let the lesser integer be x, then the next consecutive odd integer is (x+2).\r
\n" ); document.write( "\n" ); document.write( "Twice the greater (2(x+2)) is (=) 13 less than three times the lesser (3(x)-13)\r
\n" ); document.write( "\n" ); document.write( "Putting it all into an equation:\r
\n" ); document.write( "\n" ); document.write( " $\"2%28x%2B2%29+=+3x+-+13\"$ Simplify and solve for x.
\n" ); document.write( " $\"2x+%2B+4+=+3x+-+13\"$ Subtract 2x from both sides.
\n" ); document.write( " $\"4+=+x+-+13\"$ Add 13 to both sides.
\n" ); document.write( " $\"x+=+17\"$ \r
\n" ); document.write( "\n" ); document.write( "The two consecutive odd integers are: 17 and 19\r
\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "Twice 19 = 38
\n" ); document.write( "13 less than three times 17 = 51 - 13 = 38 \n" ); document.write( "

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