document.write( "Question 350814: problem 1: A boy has P 116 annual income from his two savings accounts bearing 3 percent and 5 percent interest. Then he has 25 percent more of the 3 percent savings and 40 percent more of the 5 percent savings, thereby increasing his annual income by P 41. Find his initial investment in each type of savings.\r
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document.write( " pls. help me solve this problem. tnx \n" );
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Algebra.Com's Answer #250776 by mananth(16946) You can put this solution on YOUR website! let investment in 3 % be x \n" ); document.write( "investment in 5% be y \n" ); document.write( "... \n" ); document.write( "0.03x+0.05y= 116 \n" ); document.write( "3x+5y=11600.............................1 \n" ); document.write( ".... \n" ); document.write( "25% more of 3% = 1.25x \n" ); document.write( "40% more of 5% = 1.40 y \n" ); document.write( ".. \n" ); document.write( "0.03*1.25x+0.05*1.40y=116+41 \n" ); document.write( "0.0375x+0.0700y= 157 \n" ); document.write( "375x+700y=1570000.......................2 \n" ); document.write( "multiply equation 1 by -140 and add it to equation 2 \n" ); document.write( "-420x-700y=-1624000 \n" ); document.write( "-45x=-54000 \n" ); document.write( "x =1200 @ 3% \n" ); document.write( "plug value of x in equation 1 \n" ); document.write( "3*1200+5y=11600 \n" ); document.write( "5y=11600-3600 \n" ); document.write( "5y=8000 \n" ); document.write( "y=1600 @5%\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "C H E C K \n" ); document.write( "0.05*1600 + 0.03*1200=116 \n" ); document.write( "aaaaaaah \n" ); document.write( " |