document.write( "Question 677729: Two investments were made totaling $10,500. Part of the $10,500 was invested at 9% and the remaining amount was invested at 12%. The annual interest from both investments was $1,140. How much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #421028 by partha_ban(41)\"\" \"About 
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Let, x be the amount invested at 9% interest.
\n" ); document.write( "(10500 - x) be the amount invested at 12% interest.
\n" ); document.write( "From I = Prt, interest from first investment = \"x+%2A+%289%2F100%29+%2A+1\"
\n" ); document.write( "= \"9x%2F100\"
\n" ); document.write( "Interest from 2nd investment = \"%2810500-x%29+%2A+%2812%2F100%29+%2A+1\"
\n" ); document.write( "= \"%2812%2A%2810500-x%29%29%2F100\"
\n" ); document.write( "By condition, \"9x%2F100+%2B+%2812%2A%2810500-x%29%29%2F100+=+1140\"
\n" ); document.write( "\"%289x+%2B+126000+-+12x%29%2F100+=+1140\"
\n" ); document.write( "9x + 126000 - 12x = 1140 * 100
\n" ); document.write( "9x - 12 x = 114000 - 126000
\n" ); document.write( "-3x = -12000
\n" ); document.write( "\"x+=+%28-12000%29%2F%28-3%29\"
\n" ); document.write( "x = 4000
\n" ); document.write( "Therefore, $4,000 has been invested at 9% interest.
\n" ); document.write( "$(10,500 - 4,000) = $6,500 has been invested at 12% interest.
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