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)![]() ![]() ![]() You can put this solution on YOUR website! .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( " |