document.write( "Question 1088596: Anne invested $3000 more at 12% interest than she did at 25% interest. For the year she earned $35 more on her 12% vinvestment than she did from the 25% investment. How much did she invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #702870 by ikleyn(52775) $\"\"$ $\"About$

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document.write( "Let x be the amount (in dollars) invested at 12% and y be the amount invested at 25%.\r\n" );
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document.write( "From the condition,\r\n" );
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document.write( "    x -     y = 3000,       (1)\r\n" );
document.write( "0.12x - 0.25y =   35.       (2)\r\n" );
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document.write( "This is the key moment in the solution called \"the setup\". You must look and carefully think on this system of equations.\r\n" );
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document.write( "If you understand its meaning (the meaning of each equation) then the rest is just a technique.\r\n" );
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document.write( "From (1), express x = y + 3000 and substitute it into (2). You will get a single equation for a single unknown y\r\n" );
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document.write( "0.12(y+3000) - 0.25y = 35,   or\r\n" );
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document.write( "0.12y + 360 - 0.25y = 35,\r\n" );
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document.write( "-0.13y = 35 - 360,\r\n" );
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document.write( "-0.13y = -325  ===>  y =  = 2500.\r\n" );
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document.write( "Answer.  Anne invested $2500 at 25% and $2500 + $300 = $5500 at 12%.\r\n" );
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\n" ); document.write( "\n" ); document.write( "You check the solution by substituting the found values into equations (1) and (2).\r
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