document.write( "Question 1093141: In planning her retirement, Liza deposits some money at 4.5% interest, with twice as much deposited at 5%. Find the deposited at each rate if the total annual interest income is $1450. \n" ); document.write( "
Algebra.Com's Answer #707742 by Theo(13342)\"\" \"About 
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let x = the amount invested at 4.5%.
\n" ); document.write( "let y = the amount invested at 5.0%.\r
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\n" ); document.write( "\n" ); document.write( "total interest is 1450.\r
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\n" ); document.write( "\n" ); document.write( "therefore .045*x + .05*y = 1450.\r
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\n" ); document.write( "\n" ); document.write( "y = 2 * x, so you can replace y with 2x and your equation becomes .045*x + .05*2*x = 1450\r
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\n" ); document.write( "\n" ); document.write( "factor out the x to get x * (.045 + .05*2) = 1450\r
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\n" ); document.write( "\n" ); document.write( "simplify and combine like terms to get .145 * x = 1450\r
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\n" ); document.write( "\n" ); document.write( "solve for x to get x = 10000\r
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\n" ); document.write( "\n" ); document.write( "2x is therefore equal to 20000.\r
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\n" ); document.write( "\n" ); document.write( "she has 10000 invested at 4.5% and 20000 invested at 5%.\r
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\n" ); document.write( "\n" ); document.write( "that's your solution.\r
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\n" ); document.write( "\n" ); document.write( ".045 * 10000 + .05 * 20000 = 1450\r
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