document.write( "Question 993309: Part of $8,000 was invested by an investment club at 10% interest and the rest at 12%. If the annual interest income from these investments is $900, how much was invested at each rate? \n" ); document.write( "

Algebra.Com's Answer #612584 by addingup(3677) $\"\"$ $\"About$

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Let's call the amount invested at 10% x and the amount at 12% y\r
\n" ); document.write( "\n" ); document.write( "x+y= 8000 And we can say that x= 8000-y
\n" ); document.write( "The income is:
\n" ); document.write( ".10x + .12y = $900
\n" ); document.write( "And we said that x= 8000-y, therefore:
\n" ); document.write( ".10(8,000 - y) + .12y = 900
\n" ); document.write( "800 - .10y+.12y= 900 Subtract 800 on both sides and add the y's
\n" ); document.write( ".02y= 100 Divide both sides by .02
\n" ); document.write( "y= 5000 This is the amount invested at 12%. And 8,000-5,000= 3,000 at 10%
\n" ); document.write( "Proof:
\n" ); document.write( "5000×.12= 600
\n" ); document.write( "3,000×.10= 300
\n" ); document.write( "600+300= 900 We have the right answer. \n" ); document.write( "

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