document.write( "Question 130: Ralph is four times as old as Frank. In 20 years, Ralph will be twice as old as Frank. How old are ralph and Frank? \n" ); document.write( "
Algebra.Com's Answer #104 by terrtwo(10)\"\" \"About 
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If we use R for Ralph and F for Frank, the initial relationship between their ages is R = 4F\r
\n" ); document.write( "\n" ); document.write( "Ralph's age in twenty years will be R + 20, or his age now plus 20 years.\r
\n" ); document.write( "\n" ); document.write( "Frank's age will likewise be F + 20 in twenty years\r
\n" ); document.write( "\n" ); document.write( "Therefore the second equation we can write is \r
\n" ); document.write( "\n" ); document.write( "--> (R + 20) = 2(F + 20)\r
\n" ); document.write( "\n" ); document.write( "If we then substitute 4F for R in the second equation
\n" ); document.write( "--> (4F + 20) = 2(F + 20)\r
\n" ); document.write( "\n" ); document.write( "We then distribute the 2:\r
\n" ); document.write( "\n" ); document.write( "--> 4F + 20 = 2F + 40
\n" ); document.write( "Subtract 20 from both sides;\r
\n" ); document.write( "\n" ); document.write( "--> 4F + 20 - 20 = 2F + 40 - 20
\n" ); document.write( "--> 4F = 2F + 20\r
\n" ); document.write( "\n" ); document.write( "Subtract 2F from both sides\r
\n" ); document.write( "\n" ); document.write( "--> 4F - 2F = 2F + 20 - 2F
\n" ); document.write( "--> 2F = 20\r
\n" ); document.write( "\n" ); document.write( "Divide both sides by 2;\r
\n" ); document.write( "\n" ); document.write( "--> 2F/2 = 20/2
\n" ); document.write( "--> F = 10\r
\n" ); document.write( "\n" ); document.write( "When we substitute 10 for F back into the original equation (R = 4F);\r
\n" ); document.write( "\n" ); document.write( "--> R = 4*10 = 40\r
\n" ); document.write( "\n" ); document.write( "Hence Frank is now 10 and Ralph is 40
\n" ); document.write( "Twenty years from now, Frank will be 30 and Ralph will be 60, or twice Frank's age\r
\n" ); document.write( "\n" ); document.write( "So the answer is F = 10 and R = 40
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