document.write( "Question 68087: Jack invested $8000, part at 10% and part at 15%. If the total interest at the end of the year is $900, how much did he invest at 10%?
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document.write( "a)$7000
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document.write( "b)$5000
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document.write( "c)$6000
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document.write( "d)$4000 \n" );
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Algebra.Com's Answer #48378 by checkley71(8403)![]() ![]() ![]() You can put this solution on YOUR website! .1x+.15(8000-x)=900 \n" ); document.write( ".1x+1200-.15x=900 \n" ); document.write( "-.05x=900-1200 \n" ); document.write( "-.05x=-300 \n" ); document.write( "x=-300/-.05 \n" ); document.write( "x=6000 invested @ 10% answer c) \n" ); document.write( " |