document.write( "Question 767176: Larry Mitchell invested part of his $34,000 advance at 6% annual interest rate and the rest at 4% annual interest rate. If his total yearly interest from both accounts was $1,540, find the amount at each rate. \n" ); document.write( "

Algebra.Com's Answer #467463 by MaartenRU(13) $\"\"$ $\"About$

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Let's call the part he invested for 6% x. Then the other part will be 34000-x. Now, 6% of x plus 4% of 34000-x has to be 1540, or $\"6x%2F100%2B%284%2834000-x%29%29%2F100=1540\"$ . So $\"6x%2B136000-4x=154000\"$ , or $\"2x=18000\"$ . So he invested 9000 dollars for 6% and 25000 dollars for 4%.\r
\n" ); document.write( "\n" ); document.write( "We can check, and it indeed holds: $\"0.06%2A9000%2B0.04%2A25000=540%2B1000=1540\"$ \n" ); document.write( "

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