document.write( "Question 388601: Sara has $60,000 to invest. She invest part at 4%, one half this amount at 3.5%, and the balance at 5%. Her total annual income from interest is $2650. Find the amount invested at each rate.\r
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Algebra.Com's Answer #274965 by mananth(16946)\"\" \"About 
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Sara has $60,000 to invest. She invest part at 4%, one half this amount at 3.5%, and the balance at 5%. Her total annual income from interest is $2650. Find the amount invested at each rate.
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\n" ); document.write( "Sara has $60,000 to invest.
\n" ); document.write( "at 4%, $ x
\n" ); document.write( "at 3.5%, $ x/2
\n" ); document.write( "at 5%. (60000-(x+x/2))= (60000-1.5x)
\n" ); document.write( "...
\n" ); document.write( "4x+3.5*x/2 + 5*(60000-1.5x)=2650*100
\n" ); document.write( "multiply by2
\n" ); document.write( "8x+3.5x+10(60000-1.5x)=2650*100*2
\n" ); document.write( "8x+3.5x+600000-15x=2650*100*2
\n" ); document.write( "-3.5x=-600000+530000
\n" ); document.write( "-3.5x=-70000
\n" ); document.write( "/-3.5
\n" ); document.write( "x=-70000/-3.5
\n" ); document.write( "...
\n" ); document.write( "x= $20,000 @ 4%
\n" ); document.write( "x/2 = $10,000 @3.5%
\n" ); document.write( "$30,000 @ 5%
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( "
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