document.write( "Question 1129424: Larry Mitchell invested part of his $38,000 advance at 6% annual simple interest and the rest at 2% annual simple interest. If this total yearly interest from both accounts was $880 find the amount incested at each rate.
\n" ); document.write( " The amount invested at 6% is $ ?
\n" ); document.write( " The amount invested at 2% is $ ? \n" ); document.write( "

Algebra.Com's Answer #746040 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"+x+\"$ = amount invested @ 6%
\n" ); document.write( " $\"+38000+-+x+\"$ = amount invested @ 2%
\n" ); document.write( "-----------------------------------------------
\n" ); document.write( " $\"+.06x+%2B+.02%2A%28+38000+-+x+%29+=+880+\"$
\n" ); document.write( " $\"+.06x+%2B+760+-+.02x+=+880+\"$
\n" ); document.write( " $\"+.04x+=+120+\"$
\n" ); document.write( " $\"+x+=+3000+\"$
\n" ); document.write( "and
\n" ); document.write( " $\"+38000+-+3000+=+35000+\"$
\n" ); document.write( "@ 6%
\n" ); document.write( "$3,000
\n" ); document.write( "@2%
\n" ); document.write( "$35,000
\n" ); document.write( "---------------
\n" ); document.write( "check:
\n" ); document.write( " $\"+.06%2A3000+%2B+.02%2A35000+=+880+\"$
\n" ); document.write( " $\"+180+%2B+700+=+880+\"$
\n" ); document.write( "OK \n" ); document.write( "

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