document.write( "Question 1065126: Larry Mitchell invested part of his $22,000 advance at 5% annual simple interest and the rest at 4% annual simple interest. If his total yearly interest from both accounts was $940, find the amount invested at each rate. \n" ); document.write( "

Algebra.Com's Answer #680264 by addingup(3677) $\"\"$ $\"About$

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An amount is invested at 5%. Let this amount be x. Thus, 0.05x
\n" ); document.write( "The rest is invested at 4%: 22,000-x is the amount invested at 4%. Put it all together:
\n" ); document.write( "0.05x+0.04(22,000-x) = 940
\n" ); document.write( "0.05x+880-0.04x = 940
\n" ); document.write( "0.01x = 60
\n" ); document.write( "x = 60/0.01 = 6,000 this is the amount invested at 5%
\n" ); document.write( "22,000-6,000 = 16,000 is the amount invested at 4%
\n" ); document.write( "-------------------------------------------------
\n" ); document.write( "Check:
\n" ); document.write( "6,000*0.05 = 300
\n" ); document.write( "16,000*0.04 =640
\n" ); document.write( ". . . . . .--------------
\n" ); document.write( "Total . . . 940 Correct
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\n" ); document.write( "John \n" ); document.write( "

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