document.write( "Question 536220: Part of $11,000 was invested at 10% and the rest at 12%. If the annual income from these investments was $1260, how much was invested at each rate?\r
\n" ); document.write( "\n" ); document.write( "Please, Mr/Mrs Tutor i\"ve tried countless times to evaluate this problem and several others similar to it. I need help. I need to learn how to answer any question like the stated one above by my-self, finals for me are approaching very quickly and I need to get the steps down and remembered. Thank you- Tray Tillmon \n" ); document.write( "
Algebra.Com's Answer #352213 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
First, let $x = the amount invested at 10%. The rest ($11,000-x) is the amount invested at 12%.
\n" ); document.write( "Now you need to express the amount earned in each case. Change the percentages to their decimal equivalents:
\n" ); document.write( "10% = 0.1
\n" ); document.write( "12% = 0.12
\n" ); document.write( "The amount earned on the $x can be expressed as: 0.1x and the amount earned on the rest can be expressed as 0.12($11,000-x). The sum of these two amounts is to equal $1260.00, so here's the equation that you'll need to solve for x.
\n" ); document.write( "0.1x+0.12(11000-x) = 1260 Simplify.
\n" ); document.write( "0.1x+1320-0.12x = 1260 Combine the x-terms.
\n" ); document.write( "-0.02x+1320 = 1260 Subtract 1320 from both sides.
\n" ); document.write( "-0.02x = -60 Finally, divide both sides by -0.02
\n" ); document.write( "x = $3,00 and $11,000-x = $8,000
\n" ); document.write( "$3,000 was invested at 10% and $8,000 was invested at 12%
\n" ); document.write( "Check:
\n" ); document.write( "0.1(3000)+0.12(8000) = 300+960 = 1260
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