document.write( "Question 69242: Question:
\n" ); document.write( "Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received $4 more as interest. How much amount did each of them invest a different rates?\r
\n" ); document.write( "\n" ); document.write( "So far, I have:
\n" ); document.write( "x(.12) + y(.10)= $130
\n" ); document.write( "x + y = $134
\n" ); document.write( "

Algebra.Com's Answer #49275 by checkley75(3666)\"\" \"About 
You can put this solution on YOUR website!
.12X+.10Y=130
\n" ); document.write( ".10X+.12Y=134 OR
\n" ); document.write( ".12X+.10Y-(.10X+.12Y)=-4
\n" ); document.write( ".12X-.10X+.10Y-.12Y=-4
\n" ); document.write( ".02X-.02Y=-4
\n" ); document.write( ".02X=.02Y-4
\n" ); document.write( "X=.02Y/.02-4/.02
\n" ); document.write( "X=Y-200 NOW SUBSTITUTE (Y-200) FOR X IN EITHER EQUATION. THUS:
\n" ); document.write( ".12(Y-200)+.10Y=130
\n" ); document.write( ".12Y-24+.10Y=130
\n" ); document.write( ".22Y=130+24
\n" ); document.write( ".22Y=154
\n" ); document.write( "Y=154/.22
\n" ); document.write( "Y=$700 INVESTED @ 10% TO GET THE $130 INTEREST.
\n" ); document.write( "X=7-200
\n" ); document.write( "X=500 INVESTED @ 12% TO GET $130 INTEREST.
\n" ); document.write( "PROOF
\n" ); document.write( ".10*500+.12*700=134
\n" ); document.write( "50+84=134
\n" ); document.write( "134=134
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