document.write( "Question 190617: # 9. Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate?\r
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Algebra.Com's Answer #143075 by jim_thompson5910(35256)\"\" \"About 
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# 9 \r
\n" ); document.write( "\n" ); document.write( "Q: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate? \r
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\n" ); document.write( "\n" ); document.write( "x = amount of money Mike invests at 8%\r
\n" ); document.write( "\n" ); document.write( "y = amount of money Mike invests at 12%\r
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\n" ); document.write( "\n" ); document.write( "Since \"Mike invested $7000 for one year.\", this means that \"x%2By=7000\"\r
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\n" ); document.write( "\n" ); document.write( "Also, because \"He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest.\", this translates to \"0.08x%2B0.12y=764\". Multiplying every term by 100 gets us \"8x%2B12y=76400\"\r
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\n" ); document.write( "\n" ); document.write( "So we have the system of equations:\r
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\n" ); document.write( "\n" ); document.write( "\"system%28x%2By=7000%2C8x%2B12y=76400%29\"\r
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\n" ); document.write( "\n" ); document.write( "Let's solve this system by substitution\r
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\n" ); document.write( "\n" ); document.write( "Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.\r
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\n" ); document.write( "\n" ); document.write( "So let's isolate y in the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"x%2By=7000\" Start with the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"y=7000-x\" Subtract \"x\" from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"y=-x%2B7000\" Rearrange the equation\r
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\n" ); document.write( "\n" ); document.write( "Since \"y=-x%2B7000\", we can now replace each \"y\" in the second equation with \"-x%2B7000\" to solve for \"x\"\r
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\n" ); document.write( "\n" ); document.write( "\"8x%2B12%2Ahighlight%28%28-x%2B7000%29%29=76400\" Plug in \"y=-x%2B7000\" into the second equation. In other words, replace each \"y\" with \"-x%2B7000\". Notice we've eliminated the \"y\" variables. So we now have a simple equation with one unknown.\r
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\n" ); document.write( "\n" ); document.write( "\"8x%2B%2812%29%28-1%29x%2B%2812%29%287000%29=76400\" Distribute \"12\" to \"-x%2B7000\"\r
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\n" ); document.write( "\n" ); document.write( "\"8x-12x%2B84000=76400\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"-4x%2B84000=76400\" Combine like terms on the left side\r
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\n" ); document.write( "\n" ); document.write( "\"-4x=76400-84000\"Subtract 84000 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"-4x=-7600\" Combine like terms on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28-7600%29%2F%28-4%29\" Divide both sides by -4 to isolate x\r
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\n" ); document.write( "\n" ); document.write( "\"x=1900\" Divide\r
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\n" ); document.write( "\n" ); document.write( "So this means that Mike invested $1,900 at 8%\r
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\n" ); document.write( "\n" ); document.write( "Since we know that \"x=1900\" we can plug it into the equation \"y=-x%2B7000\" (remember we previously solved for \"y\" in the first equation).\r
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\n" ); document.write( "\n" ); document.write( "\"y=-x%2B7000\" Start with the equation where \"y\" was previously isolated.\r
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\n" ); document.write( "\n" ); document.write( "\"y=-%281900%29%2B7000\" Plug in \"x=1900\"\r
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\n" ); document.write( "\n" ); document.write( "\"y=5100\" Combine like terms \r
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\n" ); document.write( "\n" ); document.write( "So Mike invested $5,100 at 12%\r
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\n" ); document.write( "\n" ); document.write( "So Mike invested $1,900 at 8% and $5,100 at 12%
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