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)\"\" \"About 
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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
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