document.write( "Question 127324: 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%?\r
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Algebra.Com's Answer #93274 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"x\"$ = the part of $31,000 invested at 10%
\n" ); document.write( "Then $\"31000+-+x\"$ will be the amount invested at 1%
\n" ); document.write( " $\".1x+%2B+.01%2831000+-+x%29+=+1390\"$
\n" ); document.write( " $\".1x+%2B+310+-+.01x+=+1390\"$
\n" ); document.write( " $\".09x+=+1080\"$
\n" ); document.write( " $\"x+=+12000\"$
\n" ); document.write( "$12,000 was invested at 10% answer
\n" ); document.write( "check:
\n" ); document.write( " $\".1x+%2B+.01%2831000+-+x%29+=+1390\"$
\n" ); document.write( " $\".1%2A12000+%2B+.01%2831000+-+12000%29+=+1390\"$
\n" ); document.write( " $\"1200+%2B+.01%2A19000+=+1390\"$
\n" ); document.write( " $\"1200+%2B+190+=+1390\"$
\n" ); document.write( " $\"1390+=+1390\"$
\n" ); document.write( "OK \n" ); document.write( "

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