document.write( "Question 99677This question is from textbook Beginning Algebra
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document.write( ": Please help me solve this equation:\r
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document.write( "Karen Waugtal invested some money at 9% annual simple interest and $250 more than that amount at 10% annual simple interest. If her total yearly interest was $101, how much was invested at each rate?\r
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document.write( "I'm stuck. \n" );
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Algebra.Com's Answer #72545 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let P=amount invested at 9%\r \n" ); document.write( "\n" ); document.write( "Then P+250=amount invested at 10%\r \n" ); document.write( "\n" ); document.write( "Interest(I)=Principal(P) times Rate(R) times Time(T) or I=PRT\r \n" ); document.write( "\n" ); document.write( "(1) Interest at 9%=P*(0.09)*(1)=0.09P\r \n" ); document.write( "\n" ); document.write( "(2) Interest at 10%=(P+250)*(0.10)*(1)=(0.10)(P+250)\r \n" ); document.write( "\n" ); document.write( "Now we are told that (1)+(2)=101 so our equation to solve is:\r \n" ); document.write( "\n" ); document.write( "0.09P+(0.10)(P+250)=101 get rid of parens\r \n" ); document.write( "\n" ); document.write( "0.09P+0.10P+25=101 subtract 25 from both sides \n" ); document.write( "0.09P+0.10P+25-25=101-25 combine like terms\r \n" ); document.write( "\n" ); document.write( "0.19P=76 divide both sides by 0.19\r \n" ); document.write( "\n" ); document.write( "P=$400------------------------------------ amount invested at 9%\r \n" ); document.write( "\n" ); document.write( "P+250=400+250=$650-----amount invested at 10%\r \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( "400*0.09+650*0.10=101\r \n" ); document.write( "\n" ); document.write( "36+65=101 \n" ); document.write( "101=101\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |