document.write( "Question 25741: A woman invested part of 20,000 at 6% and the rest at 7%. If her annual interest is 1,260, how much did she invest at 6%?\r
\n" ); document.write( "\n" ); document.write( "Interest problems always stump me. Any help with this could really help me. \n" ); document.write( "
Algebra.Com's Answer #13855 by elima(1433)\"\" \"About 
You can put this solution on YOUR website!
The formula for interest;
\n" ); document.write( "I=prt
\n" ); document.write( "I=interest
\n" ); document.write( "p=principal
\n" ); document.write( "r=rate
\n" ); document.write( "t=time
\n" ); document.write( "So you are given everything except p;
\n" ); document.write( "I=1260
\n" ); document.write( "r=7%
\n" ); document.write( "t=1
\n" ); document.write( "So, with those amounts we solve for p;
\n" ); document.write( "1260=p(.07)1
\n" ); document.write( "multiply (.07)1;
\n" ); document.write( "1260=p(.07)
\n" ); document.write( "Now divide each side by (.07)
\n" ); document.write( "\"1260%2F.07\"=\"p%28.07%29%2F.07\"
\n" ); document.write( "18,000=p
\n" ); document.write( "So $18,000 was invested at 7%, so $20,000-$18,000=$2,000
\n" ); document.write( "at 6% $2,000 was invested
\n" ); document.write( "Hope you understand
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