document.write( "Question 1178593: Allison is seven less than three times the age of Jennifer. Six years ago, one-fourth of the age of Allison is equal to the age of Jennifer was then. Find their present ages. \n" ); document.write( "
Algebra.Com's Answer #807909 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Allison age = a \n" ); document.write( "jennifer age =j\r \n" ); document.write( "\n" ); document.write( "Allison is seven less than three times the age of Jennifer. \r \n" ); document.write( "\n" ); document.write( "a=3j-7 \n" ); document.write( "a-3j =-7 \n" ); document.write( "Six years ago, one-fourth of the age of Allison is equal to the age of Jennifer \n" ); document.write( "was then.\r \n" ); document.write( "\n" ); document.write( "1/4 (a-6) = j-6 \r \n" ); document.write( "\n" ); document.write( "a-6 = 4j-24\r \n" ); document.write( "\n" ); document.write( "a-4j =-18 \n" ); document.write( "a-3j=-7\r \n" ); document.write( "\n" ); document.write( "subtract\r \n" ); document.write( "\n" ); document.write( "-j=-11\r \n" ); document.write( "\n" ); document.write( "Jennifer's age =11 years \n" ); document.write( "Allison's age ---> a=3j-7 \n" ); document.write( "=26 years\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |