document.write( "Question 747156: The age of a mother is twice that of her elder daughter. Ten years from now the age of the
\n" ); document.write( "mother will be thrice of her younger daughter. If the difference in ages of the two daughters is
\n" ); document.write( "15, the age of the mother is:
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Algebra.Com's Answer #454738 by timvanswearingen(106)\"\" \"About 
You can put this solution on YOUR website!
Sorry I misread the question the first time. \r
\n" ); document.write( "\n" ); document.write( "Let x represent the mother's age. \r
\n" ); document.write( "\n" ); document.write( "Then the elder daughter would be x/2\r
\n" ); document.write( "\n" ); document.write( "In ten years, the younger daughter's age can be represented by (x+10)/3; subtracting ten from this would give her current age: ((x+10)/3)-10\r
\n" ); document.write( "\n" ); document.write( "Since the daughters are 15 years apart, subtract the younger daughter's age from the elder's age and set it equal to 15.\r
\n" ); document.write( "\n" ); document.write( "\"x%2F2-%28%28x%2B10%29%2F3-10%29=15\"\r
\n" ); document.write( "\n" ); document.write( "Now multiply by the LCD which is 6 (2*3)\r
\n" ); document.write( "\n" ); document.write( "\"6%28x%2F2-%28%28x%2B10%29%2F3-10%29%29=15%2A6\"\r
\n" ); document.write( "\n" ); document.write( "\"3x-2%28x%2B10%29-60=90\"\r
\n" ); document.write( "\n" ); document.write( "\"3x-2x-20%2B60=90\"\r
\n" ); document.write( "\n" ); document.write( "\"x%2B40=90\"\r
\n" ); document.write( "\n" ); document.write( "\"x=50\"\r
\n" ); document.write( "\n" ); document.write( "X represented the mother's age, so the mother is 50 years old.
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