document.write( "Question 459256: a person invested 50000 into two accounts that pays 8 percent and 6 percent interest anually. find out the amount acount if the total interest earned after one year is 3600. \n" ); document.write( "

Algebra.Com's Answer #314968 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"a\"$ = amount invested @ 8%
\n" ); document.write( "Let $\"b\"$ = amount invested @ 6%
\n" ); document.write( "given:
\n" ); document.write( "(1) $\"+.08a+%2B+.06b+=+3600+\"$
\n" ); document.write( "(1) $\"+8a+%2B+6b+=+360000+\"$
\n" ); document.write( "and
\n" ); document.write( "(2) $\"+a+%2B+b+=+50000+\"$
\n" ); document.write( "Multiply both sides of (2) by $\"6\"$
\n" ); document.write( "and subtract (2) from (1)
\n" ); document.write( "(1) $\"+8a+%2B+6b+=+360000+\"$
\n" ); document.write( "(2) $\"+-6a+-+6b+=+-300000+\"$
\n" ); document.write( " $\"+2a+=+60000+\"$
\n" ); document.write( " $\"+a+=+30000+\"$
\n" ); document.write( "and, since
\n" ); document.write( "(2) $\"+a+%2B+b+=+50000+\"$
\n" ); document.write( " $\"+b+=+20000+\"$
\n" ); document.write( "$30,000 = amount invested @ 8%
\n" ); document.write( "$20,000 = amount invested @ 6%
\n" ); document.write( "check answer:
\n" ); document.write( "(1) $\"+.08a+%2B+.06b+=+3600+\"$
\n" ); document.write( "(1) $\"+.08%2A30000+%2B+.06%2A20000+=+3600+\"$
\n" ); document.write( "(1) $\"+2400+%2B+1200+=+3600+\"$
\n" ); document.write( "(1) $\"+3600+=+3600+\"$
\n" ); document.write( "OK\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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