document.write( "Question 106454: Janet invested $31,000, part at 10% and part at 1%. If the total interest at the end of the year is $1,390, how much did she invest at 10%? \n" ); document.write( "
Algebra.Com's Answer #77514 by checkley75(3666)\"\" \"About 
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.10X+.01(31,000-X)=1.390
\n" ); document.write( ".10X+310-.01X=1,390
\n" ); document.write( ".09X=1,390-310
\n" ); document.write( ".09X=1,080
\n" ); document.write( "X=1,080/.09
\n" ); document.write( "X=12,000 INVESTED @ 10%.
\n" ); document.write( "PROOF
\n" ); document.write( ".10*12,000+.01*19,000=1,390
\n" ); document.write( "1200+190=1,390
\n" ); document.write( "1,390=1,390
\n" ); document.write( "
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