document.write( "Question 232133: calvin invested $7500 for one year, part at 12% annual interest and the rest at 10% annual interest. His total for the year was $890. How much money did he invested at 12%? \n" ); document.write( "

Algebra.Com's Answer #171624 by Earlsdon(6294) $\"\"$ $\"About$

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Let x = the amount invested at 12% per annum, then ($7500-x) would be the amount invested at 10% per annum.
\n" ); document.write( "The interest, whose sum is $890, earned on these two amounts can be expressed algebraically as:
\n" ); document.write( " $\"x%280.12%29%2B+%287500-x%29%280.1%29+=+890\"$ Simplify and solve for x.
\n" ); document.write( " $\"0.12x%2B750-0.1x+=+890\"$ Combine like-terms on the left.
\n" ); document.write( " $\"0.02x%2B750+=+890\"$ Subtracr 750 from both sides.
\n" ); document.write( " $\"0.02x+=+140\"$ Divide both sides by 0.02
\n" ); document.write( " $\"x+=+7000\"$
\n" ); document.write( "The amount invested at 12% is $7,000.00 \n" ); document.write( "

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