SOLUTION: I am having a hard time setting up two equations for this word problem: A shopkeeper has two types of coffee beans. The first type sells for $5.20/lb, the second for $5.80. How

Question 127521This question is from textbook College Algebra
: I am having a hard time setting up two equations for this word problem: A shopkeeper has two types of coffee beans. The first type sells for $5.20/lb, the second for $5.80. How many pounds of the first type must be mixed with 5 lb of the second to produce a blend selling for $5.35/lb?
I have attempted setting up on equation $5.20(a) + $5.80(y) = $5.35 (1), and several other ways, and I just cannot seem to set this up algebraically. Can you please help me set up the equations so that I can attempt to solve this problem? This question is from textbook College Algebra

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
The problem requires two equations to solve.
.
The first equation will involve the total amount of mixed coffee. The total amount of mixed coffee
(call it T for total) will consist of X, the unknown number of lbs of the $5.20/lb coffee, and
5 lbs of the $5.80/lb coffee. So we can say that T, the weight of the blend is the sum of X
and 5 lbs and in equation form this is:
.
T = X + 5
.
Then we can determine the cost. We know that the cost of the blend is to be $5.35 per lb. Since
we have T lbs of the blend, the value of the blend is $5.35 per lb times T, the number of
lbs of the blend ... or 5.35 times T ... or just 5.35T.
.
The cost of the 5 lbs of $5.80/lb coffee is just the product of $5.80 times 5 lbs or
5.80 times 5 which multiplies out to $29.00.
.
And finally the cost of the X lbs of coffee at $5.20/lb is the product of these two or 5.20X.
.
Then the dollar value of the blend must equal the dollar value of the two types of coffee in the blend.
In equation form this is:
.
5.35T = 5.20X + 29.00
.
Then we can go back to the first equation T = X + 5. Since we know that T is equal to
X + 5 we can substitute X + 5 for T in the cost equation and get:
.
5.35(X + 5) = 5.20X + 29
.
Multiply out the left side by multiplying 5.35 times each of the terms in the parentheses to
get:
.
5.35X + 5.35*5 = 5.20X + 29.00
.
The product of 5 and 5.35 is 26.75. So the equation becomes:
.
5.35X + 26.75 = 5.20X + 29.00
.
Get rid of the 26.75 on the left side by subtracting 26.75 from both sides. This subtraction
changes the equation to:
.
5.35X = 5.20X + 2.25
.
Get rid of the 5.20X on the right side by subtracting 5.20X from both sides and the equation
reduces to:
.
0.15X = 2.25
.
Finally solve for X by dividing both sides by 0.15 to get:
.
X = 2.25/0.15 = 15
.
This tells you that if you mix 15 lbs of coffee at $5.20 per lb with 5 lbs of coffee at
$5.80 per lb you should get a blend that retails for $5.35 per lb.
.
Let's check that. We mix in 15 lbs of cheap coffee with 5 lbs of expensive coffee and
so we get 20 lbs of mixed coffee. The 20 lbs of mixture is supposed to be worth $5.35 per lb
and since we have 20 pounds of that blend the value of the coffee will be $5.35 times 20
which, when you multiply it out is $107.00.
.
How much was the coffee that we mixed worth? First we took 15 lbs of coffee that was worth
$5.20 per lb. So the worth of this coffee was 15 times $5.20 which multiplies out to
$78.00. To that we added 5 lbs of coffee that was worth $5.80 per lb. That multiplies
out to be $5.80 times 5 or $29.00. Therefore, the two coffees that we mixed were worth a total
of $78.00 + $29.00 = $107.00 which is the same amount of worth that we figured for the blended
mix. So our answer checks. The unknown number of lbs of the coffee at $5.20 per lb is 15 lbs.
.
Hope this helps you to understand how to develop the formulas and how to think your way through
the problem. The two equations you needed were an equation for total weight of the mix as
being the sum of the individual weights of the two coffees. And an equation that tells you
the weight of the mixed coffee at its cost per pound equals the sum of the number of
pounds of each type of coffee times the associated cost of each type.
.

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