document.write( "Question 344643: If Ben invests $4000 at 3% interest per year, how much additional money must he invest at 4.5% annual interest to ensure that the interest he receives each year is 3.5% of the total amount invested? \n" ); document.write( "
Algebra.Com's Answer #246582 by mananth(16946)\"\" \"About 
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$4000 @ 3 % Interest = $120
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\n" ); document.write( "Let him invest $ x @ 4.5%
\n" ); document.write( "interest he will get on this amount = x*0.045*1
\n" ); document.write( "=0.045x
\n" ); document.write( "..
\n" ); document.write( "Total interest will be 120 + 0.045x.............1
\n" ); document.write( "....
\n" ); document.write( "Total amount invested = 4000+x
\n" ); document.write( "interested desired = 3.5%
\n" ); document.write( "= 0.035*(4000+x ) ........................2
\n" ); document.write( "...
\n" ); document.write( "equate 1 &2
\n" ); document.write( "120+0.045x =0.035(4000+x)
\n" ); document.write( "120+0.045x=140+0.035x
\n" ); document.write( "0.045x-0.035x=140-120
\n" ); document.write( "0.01x= 20
\n" ); document.write( "/0.01
\n" ); document.write( "x=20/0.01
\n" ); document.write( "x= $2000 . investment @ 4.5%
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