document.write( "Question 37448: Vern has investments totaling $11,000 in two accounts. Once account is a savings account paying 6% interest and the other is a bond paying 9%. If the annual interest from the two investments was $855: a. How much did he have invested at the 6% rate?
\n" ); document.write( "A) $4,500 B) $6,500 C) $6,600 D) $9,900\r
\n" ); document.write( "\n" ); document.write( "b. How much did he have invested at the 9% rate?
\n" ); document.write( "A) $4,500 B) $6,500 C) $6,600 D) $9,900\r
\n" ); document.write( "\n" ); document.write( "c. How much interest did the 6% account earn?
\n" ); document.write( "A) $270 B) $585 C) $660 D) $990\r
\n" ); document.write( "\n" ); document.write( "d. How much interest did the 9% account earn?
\n" ); document.write( "A) $270 B) $585 C) $660 D) $990\r
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Algebra.Com's Answer #23084 by checkley71(8403)\"\" \"About 
You can put this solution on YOUR website!
.06X+.09(11000-X)=855 OR .06X+990-.09X=855 OR -.03X=855-990 OR .03X=135 OR \r
\n" ); document.write( "\n" ); document.write( "X=135/.03 OR X=4500 & 11000-4500=6500. 4500 INVESTED @ 6% & 6500 INVESTED @9% \r
\n" ); document.write( "\n" ); document.write( ".06*4500=270 INTEREST FOR THE 6% INVESTMENT\r
\n" ); document.write( "\n" ); document.write( ".09*6500=585 FOR THE 9% INVESTMENT
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