document.write( "Question 1196477: A woman invest $10 000;part at 9% annual interests, and the rest at 14%. In each case the interest is compounded annually. The total annual income is $1275. How much is invested in each rate? \n" ); document.write( "
Algebra.Com's Answer #829310 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "(1) by standard algebra....

\n" ); document.write( "x = amount invested at 9%
\n" ); document.write( "10000-x = amount invested at 14%

\n" ); document.write( "The total interest was $1275:

\n" ); document.write( "\".09%28x%29%2B.14%2810000-x%29+=+1275\"
\n" ); document.write( "\".09x%2B1400-.14x+=+1275\"
\n" ); document.write( "\"-.05x+=+-125\"
\n" ); document.write( "\"x+=+-125%2F-.05+=+2500\"

\n" ); document.write( "ANSWER: x = $2500 at 9%; 10000-x = $7500 at 14%

\n" ); document.write( "(2) using logical reasoning and mental arithmetic....

\n" ); document.write( "All $10,000 invested at 9% would have yielded $900 interest; all at 14% would have yielded $1400 interest.
\n" ); document.write( "Look at the three interest amounts $900, $1275, and $1400 (on a number line, if it helps) and observe/calculate that $1275 is 3/4 of the way from $900 to $1400 (1275-900 = 375; 1400-900 = 500; 375/500 = 3/4).
\n" ); document.write( "That means 3/4 of the $10,000 was invested at the higher rate.

\n" ); document.write( "ANSWER: $7500 at 14%, the other $2500 at 9%

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