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: \n" );
document.write( "
Algebra.Com's Answer #454738 by timvanswearingen(106)![]() ![]() 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( " \n" ); document.write( "\n" ); document.write( "Now multiply by the LCD which is 6 (2*3)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "X represented the mother's age, so the mother is 50 years old. \n" ); document.write( " |