document.write( "Question 997702: A broker has twice as much money invested at 8% as he has invested a 6.5% if his total income from the two investments is $2700 how much does he have invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #615673 by addingup(3677)\"\" \"About 
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Let's call the amount invested at 6.5% x.
\n" ); document.write( "The amount invested at 8% is 2x
\n" ); document.write( "-------------------
\n" ); document.write( ".065x + .08(2x)= 2,700
\n" ); document.write( ".065x + .16x= 2,700
\n" ); document.write( ".225x= 2,700 divide both sides by .225
\n" ); document.write( "x= 12,000 is what he has invested at 6.5%, and
\n" ); document.write( "12,000*2= 24,000 is the amount invested at 8%
\n" ); document.write( "Proof:
\n" ); document.write( "12,000 x 0.065= 780
\n" ); document.write( "24,000 x 0.08= 1,920
\n" ); document.write( "- - - - - - - --------
\n" ); document.write( "Total............2,700 We have the correct answer
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