document.write( "Question 927571: A total of $5000 is invested: part at 7% and the remainder at 12%. How much is invested at each rate if the annual interest is $400? \n" ); document.write( "

Algebra.Com's Answer #563077 by srinivas.g(540) $\"\"$ $\"About$

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Let x be the amount invested in an account that paid 7 % per year
\n" ); document.write( " Let y be the amount invested in an account that paid 12 % per year
\n" ); document.write( " x+y= $ 5000.......................eq(1)
\n" ); document.write( "interest on amount = 10 % of X
\n" ); document.write( " = $\"+%287%2F100%298x\"$
\n" ); document.write( " {{0.07x}}}
\n" ); document.write( "Interest on amount y = 12 % of y
\n" ); document.write( " = $\"+%2812%2F100%29%2Ay\"$
\n" ); document.write( " = $\"0.12y\"$
\n" ); document.write( "Total interest for whole amount = $ 400
\n" ); document.write( "hence
\n" ); document.write( " 0.07 x+0.12 y =400.............eq(2)
\n" ); document.write( "solve eq(1) and eq(2) to get x& Y values\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "

Solved by pluggable solver: SOLVE linear system by SUBSTITUTION

Solve:
\n" ); document.write( "

We'll use substitution. After moving 1*y to the right, we get:
\n" ); document.write( " $\"1%2Ax+=+5000+-+1%2Ay\"$ , or $\"x+=+5000%2F1+-+1%2Ay%2F1\"$ . Substitute that
\n" ); document.write( " into another equation:
\n" ); document.write( " $\"0.07%2A%285000%2F1+-+1%2Ay%2F1%29+%2B+0.12%5Cy+=+400\"$ and simplify: So, we know that y=999.999999999999. Since $\"x+=+5000%2F1+-+1%2Ay%2F1\"$ , x=4000.
\n" ); document.write( "
\n" ); document.write( " Answer: $\"system%28+x=4000%2C+y=999.999999999999+%29\"$ .
\n" ); document.write( "

\n" ); document.write( "\n" ); document.write( "Result\":\r
\n" ); document.write( "\n" ); document.write( "X= $4000
\n" ); document.write( "y= $1000 \n" ); document.write( "

\n" );