document.write( "Question 306785: you invested 4000 in 2 stocks. one is 5% and the other 7% annual interest. At the end of the year total interest was 230. how much was invested at each rate? \n" ); document.write( "

Algebra.Com's Answer #219475 by texttutoring(324) $\"\"$ $\"About$

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Let x=amount invested at 5%
\n" ); document.write( "Let y=amount invested at 7%\r
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\n" ); document.write( "\n" ); document.write( "Equation 1: x+y=4000
\n" ); document.write( "Equation 2: 0.05x + 0.07y=230\r
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\n" ); document.write( "\n" ); document.write( "Isolate for y in Eqn. 1 and substitute it into Eqn 2:\r
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\n" ); document.write( "\n" ); document.write( "y=4000-x\r
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\n" ); document.write( "\n" ); document.write( "Put this value of y into Eqn 2 instead of y:\r
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\n" ); document.write( "\n" ); document.write( "0.05x+0.07(4000-x) = 230
\n" ); document.write( "0.05x +280-0.07x = 230
\n" ); document.write( "-0.02x = -50
\n" ); document.write( "x=2500\r
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\n" ); document.write( "\n" ); document.write( "Plug this value of x into Eqn 1 to solve for y:\r
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\n" ); document.write( "\n" ); document.write( "x+y=4000
\n" ); document.write( "y=4000-x
\n" ); document.write( "y=4000-2500
\n" ); document.write( "y=1500\r
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\n" ); document.write( "\n" ); document.write( "So $2500 was invested at 5%, and $1500 was invested at 7%. \n" ); document.write( "

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