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Question 164860: I am a confused about how to set this problem up and then solve. Can someone please help me. Thank you.
You are buying coffee beans for your new business, “The Algebra Café.” You discover that low grade beans sell for $0.80 a pound, and high grade beans sell for $1.45 a pound. To meet the demands of the upcoming month, you need to buy a total of 5000 pounds of beans. You must spend the remainder of your bean money… $6, 500.
Let “x” be the number of pounds of low grade beans and let “y” be the number of pounds of high grade beans that you buy.
Write a system of linear equations (two) that describe the situation above and solve for the weight of each grade bean that you need to buy.
Thank you
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! let x = number of pounds of low grade beans.
let y = number of pounds of high grade beans.
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x + y = 5000
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low grade beans cost $.80 per pound.
high grade beans cost $1.45 per pound.
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you have $6500 to spend.
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.80 * x + 1.45 * y = 6500
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your two simultaneous equations that need to be solved are:
x + y = 5000 (first equation)
.8*x + 1.45*y = 6500 (second equation)
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multiply second equation by 1.25
two equations now look like:
x + y = 5000
x + 1.8125*y = 8125
subtract second equation from first equation.
.8125*y = 3125
divide both sides by .8125
y = 3846.153846
substitute for y in first equation.
x + 3846.153846 = 5000
x = 5000 - 3846.153846
x = 1153.846154
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substitute for x and y in second equation.
.8*x + 1.45*y = 6500
.8*1153.846154 + 1.45*3846.153846 = 6500
6500 = 6500.
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answer rounded to the nearest hundredth is:
x = number of pounds of cheap coffee = 1153.85 pounds
y = number of pounds of more expensive coffee = 3846.15 pounds.
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i assumed they meant that the coffee beans cost .80 per pound and 1.45 per pound, rathen than sold for that price. then the fact that you had 6500 to spend made more sense.
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