document.write( "Question 346808: rogers age is 1 1/3 times stevens age.eight years ago rogans age was twice stevens age.find rogers age \n" ); document.write( "
Algebra.Com's Answer #247993 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let x=Steven's age; x-8=Stevens age 8 years ago \n" ); document.write( "Then (4/3)x=Roger's age; (4/3)x-8=Roger's age 8 years ago \n" ); document.write( "Now we are told that: \n" ); document.write( "(4/3)x-8=2(x-8) multiply each term by 3 \n" ); document.write( "4x-24=6(x-8) get rid of parens \n" ); document.write( "4x-24=6x-48 add 24 to and subtract 6x from each side \n" ); document.write( "4x-6x-24+24=6x-6x+24-48 collect like terms \n" ); document.write( "-2x=-24 divide each side by -2 \n" ); document.write( "x=12-------Steven's age \n" ); document.write( "(4/3)x=(4/3)*12=16 Roger's age\r \n" ); document.write( "\n" ); document.write( "Ck \n" ); document.write( " 8 years ago, Roger was twice Steven's age: \n" ); document.write( "2(12-8)=16-8 \n" ); document.write( "8=8\r \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |