document.write( "Question 25256: Zach invests $4000, part of it at 10% annual interest and the rest at 12% annual interest. If he earned $460 in interest at the end of one year, how much did he invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #13656 by elima(1433)\"\" \"About 
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Let x be the amount invested at 10%;
\n" ); document.write( "SO we have;
\n" ); document.write( ".10x=amount invested at 10%
\n" ); document.write( "(4000-x).12=amount invested at 12%
\n" ); document.write( "So our equation would be;
\n" ); document.write( ".10x+(4000-x).12=460
\n" ); document.write( "Distribute the 12%;
\n" ); document.write( ".10x+480-.12x=460
\n" ); document.write( "Put like terms together so you can see them better;
\n" ); document.write( ".10x-.12x+480=460
\n" ); document.write( "-.02x+480=460;
\n" ); document.write( "subtract 480 from both sides;
\n" ); document.write( "-.02x=-20
\n" ); document.write( "Multiply each side by -.02;
\n" ); document.write( "x=1000
\n" ); document.write( "Now that we know what was invested at 10% we can subtract that from $4000 to get what was invested at 12%;
\n" ); document.write( "$4000-$1000=$3000
\n" ); document.write( "Hope you understand
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