document.write( "Question 751452: Currently Doris is one year older than twice Arthurs age. In ten years time the sum of their ages will be 39. Determine their current ages now. \n" ); document.write( "

Algebra.Com's Answer #457250 by kmadison(20) $\"\"$ $\"About$

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Pick all numbers that sum to 39: (39+0),(38+1),(37+2),(36+3),(35+4),(34+5),(33+6),(32+7),(31+8),(30+9),(29+10),(28+11),(27+12),(26+13),(25+14), (24+15),(23+16),(22+17),(21+18),(20+19). We can stop here because all other pairs will repeat.\r
\n" ); document.write( "\n" ); document.write( "Now remembering that we are subtracting 10 from each, we can reasonably eliminate anything 29+ since a negative number is unreasonable for an age, and an age of 0 would indicate 1 of them has not yet been born. This leaves (28+11),(27+12),(26+13),(25+14), (24+15),(23+16),(22+17),(21+18),(20+19) as possible contenders. \r
\n" ); document.write( "\n" ); document.write( "This is where we look at the 1st condition: Doris is currently 2x+1 years older than Arthur. Subtracting 10 from each pair gives us respectively: (18,1);(17,2);(16,3);(15,4);(14,5);(13,6);(12,7);(11,8);(10,9). From here we plug each pair into our equation 2x+1 and find only 1 pair satisfies the equation: (13,6) $\"2%286%29%2B1+=+13\"$ \r
\n" ); document.write( "\n" ); document.write( "So the final answer is: Doris is currently 13, Arthur is currently 6. \n" ); document.write( "

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