document.write( "Question 393387:  loan= $500,000
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document.write( "part A = 6% each year
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document.write( "part B = 11% each year\r
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document.write( "if the bank recieves $43,750 in interest each year, how much was loaned at 6%? \n" );
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| Algebra.Com's Answer #279193 by stanbon(75887)      You can put this solution on YOUR website! loan= $500,000 \n" ); document.write( "part A = 6% each year \n" ); document.write( "part B = 11% each year \n" ); document.write( "if the bank recieves $43,750 in interest each year, how much was loaned at 6%? \n" ); document.write( "------------------------ \n" ); document.write( "Equations: \n" ); document.write( "Quantity Eq::: x + y = 500,000 \n" ); document.write( "--- \n" ); document.write( "Interest Eq::0.11x + 0.06y = 43,750 \n" ); document.write( "==================== \n" ); document.write( "Multiply thru the 1st by 11. \n" ); document.write( "Multiply thru the 2nd by 100. \n" ); document.write( "-------------------------------- \n" ); document.write( "11x + 11y = 11*500,000 \n" ); document.write( "11x + 6y = 4375000 \n" ); document.write( "----------------------------- \n" ); document.write( "Subtract the 2nd from the 1st and solve for \"y\": \n" ); document.write( "5y = 1,125,000 \n" ); document.write( "--- \n" ); document.write( "y = $225,000 (amt. invested at 6%) \n" ); document.write( "================= \n" ); document.write( "500,000- 225,000 = 275,000 (amt. invested at 11% \n" ); document.write( "=============================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "============= \n" ); document.write( " |