document.write( "Question 374126: Remember Little Johnny? He now has two investments that produce a $150 income each month. If $1,000 more is invested at 9% than at 10% per year, how much was invested at each percent? There were 12 months in that year. \n" ); document.write( "

Algebra.Com's Answer #266264 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"a\"$ = amount invested at 9%
\n" ); document.write( "Let $\"b\"$ = mount invested at 10%
\n" ); document.write( "given:
\n" ); document.write( " $\"150%2A12+=+1800\"$ interest received during
\n" ); document.write( "the year from both investments
\n" ); document.write( "(1) $\"a+=+b+%2B+1000\"$
\n" ); document.write( "(2) $\".09a+%2B+.1b+=+1800\"$
\n" ); document.write( "------------------
\n" ); document.write( "Multiply both sides of (2) by $\"100\"$
\n" ); document.write( "(2) $\".09a+%2B+.1b+=+1800\"$
\n" ); document.write( "(2) $\"9a+%2B+10b+=+180000\"$
\n" ); document.write( "By substitution:
\n" ); document.write( "(2) $\"9%2A%28b+%2B+1000%29+%2B+10b+=+180000\"$
\n" ); document.write( " $\"9b+%2B+9000+%2B+10b+=+180000\"$
\n" ); document.write( " $\"19b+=+180000+-+9000\"$
\n" ); document.write( " $\"19b+=+171000\"$
\n" ); document.write( " $\"b+=+9000\"$
\n" ); document.write( "And, from (1)
\n" ); document.write( " $\"a+=+9000+%2B+1000\"$
\n" ); document.write( " $\"a+=+10000\"$
\n" ); document.write( "$10,000 was invested at 9% and
\n" ); document.write( "$9,000 was invested at 10% \n" ); document.write( "

\n" );