document.write( "Question 965809: Dave has $2,000 to invest. He would like to earn $$135.20 per year in interest. How much should he invest at 6% if the rest is to be invested at 7%?
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Algebra.Com's Answer #590308 by macston(5194)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( "X=amount at 6%; Y=amount at 7% \n" ); document.write( "X+Y=$2000 Total amount invested is $2000 \n" ); document.write( "X=$2000-Y Use this to substitute for X \n" ); document.write( ". \n" ); document.write( "0.06X+0.07Y=$135.20 (Amount of interest at both rates) Substitute for X from above. \n" ); document.write( "0.06($2000-Y)+0.07Y=$135.20 \n" ); document.write( "$120-0.06Y+0.07Y=$135.20 Subtract $120 from each side. \n" ); document.write( "0.01Y=$15.20 Multiply each side by 100 \n" ); document.write( "Y=$1520.00 He should invest $1520 at 7%. \n" ); document.write( "X=$2000-Y=$2000-$1520=$480 ANSWER: He should invest $480 at 6%. \n" ); document.write( ". \n" ); document.write( "CHECK: \n" ); document.write( "0.06X+0.07Y=$135.20 \n" ); document.write( "0.06($480)+0.07($1520)=$135.20 \n" ); document.write( "$28.80+$106.40=$135.20 \n" ); document.write( "$135.20=$135.20 \n" ); document.write( " |