document.write( "Question 1045603: A self-emloyed contractor nearing retiremengt made two investments totaling $15,000. In one year, these investements yielded $573 in 3% simple interest. paty of the money was investedat 3% and theorest at 4.5%. How much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #661083 by garettbuff(18)\"\" \"About 
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Set up the problem .03 (15000-x)+.045=573
\n" ); document.write( "Simplify 450-.03x+.045x=573
\n" ); document.write( "Simplify .015x+450=573
\n" ); document.write( "Simplify .015=123
\n" ); document.write( "Multiply by 1000
\n" ); document.write( "15x=123000
\n" ); document.write( "Simplify
\n" ); document.write( "123000÷15=8,200=x
\n" ); document.write( "15000-8200=6800\r
\n" ); document.write( "\n" ); document.write( "So 3% at $6,800
\n" ); document.write( "And 4.5% at $8,200
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