document.write( "Question 123352: 8. Business and finance. A coffee merchant has coffee beans that sell for $9 per pound and $12 per pound. The two types are to be mixed to create 100 lb of a mixture that will sell for $11.25 per pound. How much of each type of bean should be used in the mixture?
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Algebra.Com's Answer #90537 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Let x= # of pounds for $9 coffee beans and y= # of pounds for $12 coffee beans \r
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\n" ); document.write( "\n" ); document.write( "Since the merchant wants to create a 100lb mixture, this means that the sum of the two types of beans is 100. So we have the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"x%2By=100\"\r
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\n" ); document.write( "\n" ); document.write( "Now since the merchant wants to mix the beans to sell at $11.25, we have the second equation\r
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\n" ); document.write( "\n" ); document.write( "\"9x%2B12y=11.25%28100%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"9x%2B12y=1125\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "So our system is \r
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\n" ); document.write( "\n" ); document.write( "\"system%28x%2By=100%2C9x%2B12y=1125%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now in order to solve this system by using elimination/addition, 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( "In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for \"y\", we would have to eliminate \"x\" (or vice versa).\r
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\n" ); document.write( "\n" ); document.write( "So lets eliminate \"x\". In order to do that, we need to have both \"x\" coefficients that are equal in magnitude but have opposite signs (for instance 2 and -2 are equal in magnitude but have opposite signs). This way they will add to zero. By adding to zero, they can be eliminated.\r
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\n" ); document.write( "\n" ); document.write( "So to make the \"x\" coefficients equal in magnitude but opposite in sign, we need to multiply both \"x\" coefficients by some number to get them to an common number. So if we wanted to get \"1\" and \"9\" to some equal number, we could try to get them to the LCM.\r
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\n" ); document.write( "\n" ); document.write( "Since the LCM of \"1\" and \"9\" is \"9\", we need to multiply both sides of the top equation by \"9\" and multiply both sides of the bottom equation by \"-1\" like this:\r
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\n" ); document.write( "\n" ); document.write( "\"9%28x%2By%29=9%28100%29\" Multiply the top equation (both sides) by \"9\"
\n" ); document.write( "\"-1%289x%2B12y%29=-1%281125%29\" Multiply the bottom equation (both sides) by \"-1\"\r
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\n" ); document.write( "\n" ); document.write( "Distribute and multiply\r
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\n" ); document.write( "\n" ); document.write( "\"9x%2B9y=900\"
\n" ); document.write( "\"-9x-12y=-1125\"\r
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\n" ); document.write( "\n" ); document.write( "Now add the equations together. In order to add 2 equations, group like terms and combine them\r
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\n" ); document.write( "\n" ); document.write( "\"%289x-9x%29%2B%289y-12y%29=900-1125\"\r
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\n" ); document.write( "\n" ); document.write( "Combine like terms and simplify\r
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\n" ); document.write( "\n" ); document.write( "\"cross%289x-9x%29-3y=-225\" Notice how the x terms cancel out\r
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\n" ); document.write( "\n" ); document.write( "\"-3y=-225\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "\"y=-225%2F-3\" Divide both sides by \"-3\" to isolate y\r
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\n" ); document.write( "\n" ); document.write( "\"y=75\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "Now plug this answer into the top equation \"x%2By=100\" to solve for x\r
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\n" ); document.write( "\n" ); document.write( "\"x%2By=100\" Start with the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"x%2B%2875%29=100\" Plug in \"y=75\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=100-75\"Subtract 75 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"x=25\" Combine like terms on the right side\r
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\n" ); document.write( "\n" ); document.write( "So our answer is\r
\n" ); document.write( "\n" ); document.write( "\"x=25\" and \"y=75\"\r
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\n" ); document.write( "\n" ); document.write( "So the merchant needs 25 pounds of $9 beans and 75 pounds of $12 beans \r
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