document.write( "Question 260969: Last year, Jodi invested $10,000, part at 6% annual interest and the rest at 8% annual interest. If she received $760 in interest at the end of the year, how much did she invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #192307 by checkley77(12844) $\"\"$ $\"About$

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.08x+.06(10,000-x)=760
\n" ); document.write( ".08x+600-.06x=760
\n" ); document.write( ".02x=760-600
\n" ); document.write( ".02x=160
\n" ); document.write( "x=160/.02
\n" ); document.write( "x=8,000 amount invested @ 8%
\n" ); document.write( "10,000-8,000=2,000 invested @ 6%
\n" ); document.write( "Proof:
\n" ); document.write( ".08*8,000+.06*2,000=760
\n" ); document.write( "640+120=760
\n" ); document.write( "760=760 \n" ); document.write( "

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