document.write( "Question 269447: Bob has some money invested at 5%, and $5,000 more than this amount invested at 9%. His annual interest income is $1430. What is the amount invested at each rate. \n" ); document.write( "

Algebra.Com's Answer #197488 by drk(1908) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let Bob have x dollars to invest. we want the equation
\n" ); document.write( "I = prt + prt
\n" ); document.write( "for two investments.
\n" ); document.write( "From above we get
\n" ); document.write( "(i) $\"1430+=+x%28.05%29%281%29+%2B+%28x%2B5000%29%2A%28.09%29%281%29\"$
\n" ); document.write( "step 1 - multiply and simplify the right side to get
\n" ); document.write( "(ii) $\"+1430+=+.05x+%2B+.09x+%2B+450\"$
\n" ); document.write( "and then
\n" ); document.write( "(iii) $\"1430+=+.14x+%2B+450\"$
\n" ); document.write( "step 2 - subtract 450 to get
\n" ); document.write( "(iv) $\"980+=+.14x\"$
\n" ); document.write( "step 3 - divide by .14 to get
\n" ); document.write( "(v) $\"x+=+7000\"$
\n" ); document.write( "and then $\"x+%2B+5000+=+12000\"$
\n" ); document.write( "-----
\n" ); document.write( "7000 @ 5%
\n" ); document.write( "12000 @ 9% \n" ); document.write( "

\n" );