document.write( "Question 201804: the father's age 10 years ago was 35 years more than twice his son's age. After how many years from now will the father be
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document.write( "(i) twice his son's age?
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document.write( "(ii) thrice his son's age? \n" );
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Algebra.Com's Answer #152157 by MathTherapy(10552)![]() ![]() You can put this solution on YOUR website! Let son’s age now be s \n" ); document.write( "Then, 10 years ago, son’s age was s – 10\r \n" ); document.write( "\n" ); document.write( "Therefore, father’s age, 10 years ago was 2(s – 10) + 35 = 2s – 20 +35 = 2s + 15, which means that father’s age now is 2s + 15 + 10 = 2s + 25\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To find the amount of years for father’s age to be twice his son’s, and using y as those amount of years, we get: 2(s + y) = 2s + 25 + y\r \n" ); document.write( "\n" ); document.write( "2s + 2y = 2s + 25 + y\r \n" ); document.write( "\n" ); document.write( "y = 25\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So, in \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To find the amount of years for father’s age to be thrice his son’s, and using y as those amount of years, we get: 3(s + y) = 2s + 25 + y\r \n" ); document.write( "\n" ); document.write( "3s + 3y = 2s + 25 + y\r \n" ); document.write( "\n" ); document.write( "2y = - s + 25\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So, in \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |