document.write( "Question 927328: $5900 is invested,part of it at 10% and part of it at 8%. For a certain year, the total yield is $534.00. How much was invested at each rate? \n" ); document.write( "

Algebra.Com's Answer #563084 by srinivas.g(540) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x be the amount invested @ 10 % per year
\n" ); document.write( " Let y be the amount invested @ 8 % per year
\n" ); document.write( " x+y= $ 5900.......................eq(1)
\n" ); document.write( "interest on amount x = 10 % of X
\n" ); document.write( " = $\"+%2810%2F100%298x\"$
\n" ); document.write( " {{0.1x}}}
\n" ); document.write( "Interest on amount y = 8 % of y
\n" ); document.write( " = $\"+%288%2F100%29%2Ay\"$
\n" ); document.write( " = $\"0.08y\"$
\n" ); document.write( "Total interest for whole amount = $ 534
\n" ); document.write( "hence
\n" ); document.write( " 0.1x x+0.08 y =534.............eq(2)
\n" ); document.write( "\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( " $\"1%2Ax+=+5900+-+1%2Ay\"$ , or $\"x+=+5900%2F1+-+1%2Ay%2F1\"$ . Substitute that
\n" ); document.write( " into another equation:
\n" ); document.write( " $\"0.1%2A%285900%2F1+-+1%2Ay%2F1%29+%2B+0.08%5Cy+=+534\"$ and simplify: So, we know that y=2800. Since $\"x+=+5900%2F1+-+1%2Ay%2F1\"$ , x=3100.
\n" ); document.write( "
\n" ); document.write( " Answer: $\"system%28+x=3100%2C+y=2800+%29\"$ .
\n" ); document.write( "

\n" ); document.write( "\n" ); document.write( "Result\": x= $ 3100
\n" ); document.write( " y=$2800
\n" ); document.write( " \n" ); document.write( "

\n" );