document.write( "Question 144619: 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 at different rates?
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Algebra.Com's Answer #105416 by checkley77(12844)\"\" \"About 
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.12X+.10Y=130 MULTIPLY THIS EQUATION BY -10
\n" ); document.write( ".10X+.12Y=134 MULTIPLY THIS EQUATION BY 12 & THRN ADD THEM.
\n" ); document.write( "-1.2X-1Y=-1300
\n" ); document.write( "1.2X+1.44Y=1608
\n" ); document.write( "-----------------------
\n" ); document.write( ".44Y=308
\n" ); document.write( "Y=308/.44
\n" ); document.write( "Y=$700 INVESTED BY JANE @ 10%.
\n" ); document.write( ".12X+.10*700=130
\n" ); document.write( ".12X+70=130
\n" ); document.write( ".12X=130-70
\n" ); document.write( ".12X=60
\n" ); document.write( "X=60/.12
\n" ); document.write( "X=500 INVESTED BY JANE @ 12%.
\n" ); document.write( "PROOF:
\n" ); document.write( ".12*500+.10*700=130
\n" ); document.write( "60+70=130
\n" ); document.write( "130=130
\n" ); document.write( "AND:
\n" ); document.write( ".10*500+.12*700=134
\n" ); document.write( "50+84=134
\n" ); document.write( "134=134\r
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