document.write( "Question 269869: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year. If the total yearly interest from this investment is $256, how much did Pat invest at 10 percent and how much at 8 percent? \n" ); document.write( "
Algebra.Com's Answer #197705 by checkley77(12844)\"\" \"About 
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.10x+.08(3,000-x)=256
\n" ); document.write( ".10x+240-.08x=256
\n" ); document.write( ".02x=256-240
\n" ); document.write( ".02x=16
\n" ); document.write( "x=16/.02
\n" ); document.write( "x=$800 amount invested @ 10%.
\n" ); document.write( "3,000-800=$2,200 amount invested @ 8%.
\n" ); document.write( "Proof:
\n" ); document.write( ".10*800+.08*2,200=256
\n" ); document.write( "80+176=256
\n" ); document.write( "256=256\r
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