document.write( "Question 593730: I am just re-learning algebra after 20 years, so it's been difficult.\r
\n" ); document.write( "\n" ); document.write( "A college student earned $7300 during summer vacation working as a waiter in a popular restaurant. The student invested part of the money at 7% and the rest at 6%. If the student received a total of $458 in interest at the end of the yera, how much was invested at 7%?\r
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Algebra.Com's Answer #376431 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
Part I 7.00% per annum ---x
\n" ); document.write( "Part II 6.00% per annum ---y
\n" ); document.write( "Sum of their investments
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\n" ); document.write( "x +y= 7300 -------------------1
\n" ); document.write( "Interest from both investments
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\n" ); document.write( "7.00% x + 6.00% y= 458
\n" ); document.write( "Multiply by 100
\n" ); document.write( "7 x + 6 y= 45800.00 --------2
\n" ); document.write( "Multiply (1) by -7
\n" ); document.write( "we get
\n" ); document.write( "-7 x -7 y= -51100.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x -1 y= -5300
\n" ); document.write( "divide by -1
\n" ); document.write( " y = 5300
\n" ); document.write( "Part I 6.00% $ 5300
\n" ); document.write( "Part II 7.00% $ 2000
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\n" ); document.write( "CHECK
\n" ); document.write( "5300 --------- 6.00% ------- 318.00
\n" ); document.write( "2000 ------------- 7.00% ------- 140.00
\n" ); document.write( "Total -------------- 458.00
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\n" ); document.write( "m.ananth@hotmail.ca
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