document.write( "Question 517629: A total of $1200 is invested, partly at 3% and the rest at 5%.Find the amounts invested at each rate if the total interest is $ 54
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Algebra.Com's Answer #344865 by Maths68(1474)\"\" \"About 
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Let
\n" ); document.write( "At 3% amount invested = x
\n" ); document.write( "At 5% amount invested = 1200-x\r
\n" ); document.write( "\n" ); document.write( "3%*x+(1200-x)*5% = 54
\n" ); document.write( "(3/100)*x+(1200-x)*(5/100)=54
\n" ); document.write( "(3x/100)+(1200*5)/100-(5x/100)=54
\n" ); document.write( "(3x/100)+(6000)/100-(5x/100)=54\r
\n" ); document.write( "\n" ); document.write( "Multiply by 100 both sides of the equation
\n" ); document.write( "3x+6000-5x=5400
\n" ); document.write( "-2x=5400-6000
\n" ); document.write( "-2x=-600
\n" ); document.write( "-2x/-2=-600/-2
\n" ); document.write( "x=300\r
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\n" ); document.write( "\n" ); document.write( "At 3% amount invested = x = $300
\n" ); document.write( "At 5% amount invested = 1200-x=1200-300=$900\r
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\n" ); document.write( "Check
\n" ); document.write( "300 * 3/100 =3*3= $9
\n" ); document.write( "900 * 5/100 = 9*5=$45
\n" ); document.write( "sum = $9+$45= $54\r
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