document.write( "Question 179560: 3. Walt made an extra $12,000 last year from a part-time job. He invested part of the money at 7% and the rest at 9%. He made a total of $970 in interest. How much was invested at 7%?\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #134505 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Let x = the amount Walt invested at 7%, so the remainder ($12000-x) was invested at 9%.
\n" ); document.write( "The interest earned on these two amounts can be expressed as:
\n" ); document.write( "(7%)(x) + (9%)($12000-x) and the sum of these two amounts is given as $970.00
\n" ); document.write( "Now you can write the equation that will allow you to so;ve for x, the amount invested at 7%.
\n" ); document.write( "First, change the percentages to their decimal equivalents:
\n" ); document.write( "(7% = 0.07) and (9% = 0.09)
\n" ); document.write( "\"0.07x%2B0.09%2812000-x%29+=+970\" Simplifying this, you'll get:
\n" ); document.write( "\"0.07x%2B1080-0.09x=+970\" Combine the x-terms.
\n" ); document.write( "\"-0.02x%2B1080+=+970\" Subtract 1080 from both sides of the equation.
\n" ); document.write( "\"-0.02x+=+-110\" Finally, divide both sides by -0.02
\n" ); document.write( "\"highlight%28x+=+5500%29\"
\n" ); document.write( "$5,500.00 was invested at 7% while $12000-$5500 = $6,500.00 was invested at 9%
\n" ); document.write( "
\n" );