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)![]() ![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " \n" ); document.write( " |