document.write( "Question 537454: John invested a portion of $15,000 at 10% annual interest and the rest at 6% annual interest. If he earned $1,260 for the year from the two accounts, how much did he invest at 10%? \n" ); document.write( "

Algebra.Com's Answer #352872 by mananth(16946) $\"\"$ $\"About$

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Investment I 6.00% per annum ---x
\n" ); document.write( "Investment II 10.00% per annum ---y
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\n" ); document.write( "x + y= 15000 ------------------------1
\n" ); document.write( "6.00% x + 10.00% y= = $1,260.00
\n" ); document.write( "Multiply by 100
\n" ); document.write( "6 x + 10 y= = $126,000.00 --------2
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\n" ); document.write( "Multiply (1) by -6
\n" ); document.write( "we get
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\n" ); document.write( "-6 x -6 y= = -90000.00
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\n" ); document.write( "Add this to (2)
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\n" ); document.write( "0 x 4 y= = $36,000.00
\n" ); document.write( "
\n" ); document.write( "divide by 4
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\n" ); document.write( "y = $9,000.00 investment at 10.00%
\n" ); document.write( "Balance $6,000.00 investment at 6.00%
\n" ); document.write( "CHECK
\n" ); document.write( "$6,000.00 --------- 6.00% ------- $360.00
\n" ); document.write( "$9,000.00 ------- 10.00% ------- $900.00
\n" ); document.write( "Total -------------- $1,260.00 \r
\n" ); document.write( "\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( "

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