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Published byMaría Mercedes Montero Casado Modified over 6 years ago
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Mixture Problems In two variables
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Mixture Problems Now that we know how to solve equations involving more than one variable, story problems will be much easier! We will use x to represent one item in the story and y to represent the other item.
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Mixture Problems A coffee shop owner wants to make 50 lbs of a new coffee blend by mixing beans that sell for $12/lb with beans that sell for $8/lb. She will sell for the new blend for $9/lb. How many pounds of each type of bean should she use?
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Mixture Problems Let x = the number of pounds of $12/lb beans
Let y = the number of pounds of $8/lb beans We know that she is making 50 pounds of the new blend so our first equation is x + y = 50
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Mixture Problems Let x = the number of pounds of $12/lb beans
The value of the $12/lb beans is $12 times the number of pounds of those beans (x). Therefore, the value of those beans is 12x
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Mixture Problems Let y = the number of pounds of $8/lb beans
The value of the $8/lb beans is $8 times the number of pounds of those beans (y). Therefore, the value of those beans is 8y.
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Mixture Problems We know the value of the new blend is
the value of the $12 beans + the value of the $8 beans or 12x + 8y. And we know that 50 pounds of beans worth $9/pound is 50(9) or $450. 12x + 8y = 450
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Now we can put our two equations together:
Mixture Problems Now we can put our two equations together: x + y = 50 12x + 8y = 450
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Mixture Problems We know how to solve that!
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Mixture Problems Let’s check it! Our original equations were:
x + y = 50 12x + 8y = 450 = 50 check 12.5(12) (8) = 450 = 450 check
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are easier when you use two variables!
Mixture Problems are easier when you use two variables!
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