Frustrating System of Linear Equations Problem

[MrTully]

I've been struggling with this problem for hours now... because I can't wrap my mind around the wording here. I'm supposed to translate this word problem into a system of linear equations:

Jimmy is a partner in an Internet-based coffee supplier. The company offers gourmet coffee beans for $12 per pound and regular coffee beans for $5 per pound. Jimmy is creating a medium-price product that will sell for $7 per pound. The first thing to go into the mixing bin was 14 pounds of the gourmet beans. How many pounds of the less expensive regular beans should be added.

Any help here would be great. Thanks in advance.

 

[stapel]

One option might be to try using the standard "mixture" formulation:

. . .units (pounds):
. . . . .gourmet: 14
. . . . .regular: r
. . . . .mixture: 14 + r

. . .rate per unit ($/lb):
. . . . .gourmet: 12
. . . . .regular: 5
. . . . .mixture: ??

. . .total (rate × units):
. . . . .gourmet: (12)(14)
. . . . .regular: (5)(r)
. . . . .mixture: ??

Complete the table.

Since the cost of the inputs must equal the cost of the output, sum the inputs and set the total equal to the output.

Solve for the variable. :wink:

Note: This can also be set up using "grids", though the reasoning is exactly the same.

Hope that helps!

Eliz.

 

[Mrspi]

I've been struggling with this problem for hours now... because I can't wrap my mind around the wording here. I'm supposed to translate this word problem into a system of linear equations:

Jimmy is a partner in an Internet-based coffee supplier. The company offers gourmet coffee beans for $12 per pound and regular coffee beans for $5 per pound. Jimmy is creating a medium-price product that will sell for $7 per pound. The first thing to go into the mixing bin was 14 pounds of the gourmet beans. How many pounds of the less expensive regular beans should be added.

Any help here would be great. Thanks in advance.

Let x = number of pounds of cheaper beans
Since each pound is worth $5, the total value of the cheaper beans is 5x

We know that he uses 14 pounds of gourmet beans. These are worth $12 a pound, so the value of the gourmet beans is 12(14).

The final mixture will contain a total of 14 + x pounds, and will be worth $7 a pound. So the value of the final mixture is 7(14 + x).

Now.....

value of gourmet beans + value of cheap beans = value of final mixture
12(14) + 5(x) = 7(14 + x)

I don't see any reason to use a system of equations for this problem....it seems easier to set up with just one variable. That's my opinion....others may differ.

(You're too fast for me, Eliz! )

 

[MrTully]

Thank you, but I'm not sure I understand. How do I figure out how many pounds of the regular beans should be in the final mixture? Should I solve for x in that equation?

I know I'm probably making myself sound like a total idiot here...

I don't see any reason to use a system of equations for this problem....it seems easier to set up with just one variable. That's my opinion....others may differ.

My instructor stated on the paper that we have to "Write and solve a system of equations in two variables for the applications." In order to get full credit on the assignment, I guess I need to set it up as a system.

 

[MrTully]

:roll: Sorry about being such a numbskull there. I think (hope) I figured out how it should look as a system, thanks to you guys:

let x = lbs of regular beans in the final mixture; let y = value of final mixture in dollars.

y = 12(14) + 5x
y = 7(14 + x)

then,

y = 168 + 5x
y = 98 + 7x

Using substitution, it comes out as (98+7x = 168+5x), which then simplifies down to (x = 35). So, there are 35 pounds of regular beans in the final mixture...

I hope that's right. May I ask if someone can confirm it for me? Thank you . And sorry once again XD.

 

[Denis]

Yepper; yer correct.

I'd set 'em up like this:

y = x + 14 [1]
7y = 5x + 14(12) [2]

Then, substitute [1] in [2]:
7(x + 14) = 5x + 14(12)
7x + 98 = 5x + 168
2x = 70
x = 35

That way, your instructor will stick a star on your forehead :wink:

 

[soroban]

Hello, MrTully!

I agree with Mrspi . . . one variable is required.
. . I'll rehash her excellent solution.

Jimmy is a partner in an Internet-based coffee supplier.
The company offers gourmet coffee beans for $12/lb and regular coffee beans for $5/lb.
Jimmy is creating a medium-price product that will sell for $7 per pound.
The first thing to go into the mixing bin was 14 pounds of the gourmet beans.
How many pounds of the less expensive regular beans should be added?

He has 14 pounds of $12 coffee.
. . Its value is: \(\displaystyle \,14\,\times\,\$12\:=\:\$168\)

He adds \(\displaystyle x\) pounds of $5 coffee.
. . Its value is: \(\displaystyle \,x \,\times\, \$5 \:=\:5x\) dollars.

Hence, the total value of the mixture is: \(\displaystyle \,5x\,+\,168\) dollars. .[1]


Let's look at the problem again.

He started with \(\displaystyle 14\) of coffee.
He added \(\displaystyle x\) pounds of coffee.
. . Hence, the mixture contains \(\displaystyle x\,+\,14\) pounds of coffee.
This mxiture is supposed to be worth $7 per pound.

Hence, the total value of the mixture is: \(\displaystyle \,7(x\,+\,14)\) dollars. .[2]


We just expressed the total value of the mixture in two ways, [1] and [2].

There is our equation! . . . . . \(\displaystyle \L\fbox{5x\,+\,168\:=\:7(x\,+\,14)}\)