document.write( "Question 902512: A woman has a total of $10,000 to invest. She invests part of the money in an account that pays 6% per year and the rest in an account that pays 10% per year. If the interest earned in the first year is $760, how much did she invest in each account? \n" ); document.write( "
Algebra.Com's Answer #547294 by richwmiller(17219)\"\" \"About 
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We know the total amount of money invested. $10000
\n" ); document.write( "x+y=10000,
\n" ); document.write( "We know that the total interest for the year for the two accounts is $760
\n" ); document.write( "0.06*x+0.1*y=760
\n" ); document.write( "x=10000-y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.06*(10000-y)+0.1*y=760
\n" ); document.write( "We multiply out
\n" ); document.write( "600-0.06y+0.1*y=760
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.04*y=160
\n" ); document.write( "Isolate y
\n" ); document.write( "y=4000 at 10%
\n" ); document.write( "x=10000-y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=6000 at 6%
\n" ); document.write( "We check
\n" ); document.write( "0.06*6000+0.1*4000=760
\n" ); document.write( "360+400=760
\n" ); document.write( "760=760\r
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