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)![]() ![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( "=) \n" ); document.write( " \n" ); document.write( " |