document.write( "Question 181917This question is from textbook algebra 1
\n" ); document.write( ": martha has $6000 to invest. she puts x dollars of this money into a savings account that earns 3% per year, and with the rest, she buys a certificate of deposit that earns 6% per year. suppose at the end of one year, martha has a total of $6315. how much money did martha invest in each account? \n" ); document.write( "
Algebra.Com's Answer #136583 by mangopeeler07(462)\"\" \"About 
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x---money invested at 3%
\n" ); document.write( "y---money invested at 6%\r
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\n" ); document.write( "\n" ); document.write( "1x+1y=6000----------b/c she has 6000 to invest
\n" ); document.write( "1.03x+1.06y=6315-------------b/c of the percent of investment and the final amount of money\r
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\n" ); document.write( "\n" ); document.write( "solve eq. 1 for x
\n" ); document.write( "x=6000-y\r
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\n" ); document.write( "\n" ); document.write( "plug that in to the second eq.
\n" ); document.write( "1.03(6000-y)+1.06y=6315\r
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\n" ); document.write( "\n" ); document.write( "Distribute
\n" ); document.write( "6180-1.03y+1.06y=6315\r
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\n" ); document.write( "\n" ); document.write( "Combine like terms
\n" ); document.write( ".03y=135\r
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\n" ); document.write( "\n" ); document.write( "Solve for y
\n" ); document.write( "y=4500\r
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\n" ); document.write( "\n" ); document.write( "x=6000-y
\n" ); document.write( "x=6000-4500
\n" ); document.write( "x=1500\r
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\n" ); document.write( "\n" ); document.write( "Check answers:
\n" ); document.write( "x+y=6000
\n" ); document.write( "1500+4500=6000-------true\r
\n" ); document.write( "\n" ); document.write( "1.03(1500)+1.06(4500)=6315-----true
\n" ); document.write( "-------------------------------------------
\n" ); document.write( "Answers:
\n" ); document.write( "money invested at 3%----$1500
\n" ); document.write( "money invested at 6%----$4500
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