document.write( "Question 1057008: An investment of Php 100 000 earned a net income of Php 3 450 in one year. Part of the investment was in bonds and earned an income at a rate of 8%. The rest of the investment was in stocks and lost money at a rate of 5%. How much was invested in bonds?
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Algebra.Com's Answer #672097 by solve_for_x(190)\"\" \"About 
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Let B represent the amount invested in bonds, and let S represent the amount invested in stocks.\r
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\n" ); document.write( "\n" ); document.write( "The sum of the two must equal 100,000, which gives:\r
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\n" ); document.write( "\n" ); document.write( "B + S = 100,000\r
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\n" ); document.write( "\n" ); document.write( "The amount of income earned is then:\r
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\n" ); document.write( "\n" ); document.write( "0.08B - 0.05S = 3450\r
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\n" ); document.write( "\n" ); document.write( "Solving the first equation for S gives:\r
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\n" ); document.write( "\n" ); document.write( "S = 100,000 - B\r
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\n" ); document.write( "\n" ); document.write( "Substituting this into the second equation gives:\r
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\n" ); document.write( "\n" ); document.write( "0.08B - 0.05(100,000 - B) = 3450\r
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\n" ); document.write( "\n" ); document.write( "0.08B + 0.05B - 5000 = 3450\r
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\n" ); document.write( "\n" ); document.write( "0.13B = 3450 + 5000\r
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\n" ); document.write( "\n" ); document.write( "0.13B = 8450\r
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\n" ); document.write( "\n" ); document.write( "B = 8450 / 0.13\r
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\n" ); document.write( "\n" ); document.write( "B = 65,000\r
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\n" ); document.write( "\n" ); document.write( "The amount invested in bonds was Php 65,000.
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