Linear Equations - Cost per Pound of Coffee Beans

[AROJ]

My problem is:
A coffee house blended 18 pounds of espresso flavored coffee beans with 17 pounds of vanilla flavored coffee beans. The 35 pound mixture cost \$306.50. A second mixture included 19 pounds of espresso flavored coffee beans and 15 pounds of vanilla flavored coffee beans. The 34 pound mixture cost \$298.50. Find the cost per pound of the espresso and vanilla flavored coffee beans

So I know,

18E+17V=306.50
19E+15V=298.50

What next?

 

[SuperSonic4]

My problem is:
A coffee house blended 18 pounds of espresso flavored coffee beans with 17 pounds of vanilla flavored coffee beans. The 35 pound mixture cost \$306.50. A second mixture included 19 pounds of espresso flavored coffee beans and 15 pounds of vanilla flavored coffee beans. The 34 pound mixture cost \$298.50. Find the cost per pound of the espresso and vanilla flavored coffee beans

So I know,

18E+17V=306.50 ... (1)
19E+15V=298.50 ... (2)

What next?

where (1) and (2) are the equations in question

You have two equations and two unknowns so you can solve using simultaneous equations. You can either do this by elimination or substitution. In this case I would use the latter. To do this make E or V the subject of an equation and substitute it's value for E or V in the other equation. In this case...

\(\displaystyle 18E + 17V = 306.50 \)

\(\displaystyle V = \dfrac{306.50-18E}{17}\)

You can now sub \(\displaystyle \dfrac{306.50-18E}{17}\) for V in (2)

\(\displaystyle 19E + 15 \left(\dfrac{306.50-18E}{17}\right) = 298.50\)

This gives an equation in E only which means you can find E and hence V

 

[MarkFL]

I did a few things here...I put a backslash in front of the dollar signs to that they would not be parsed as $\LaTeX$ tags, I edited your first equation so that the leading coefficient is 18, and I moved your post to its own thread. We prefer that new questions be posted in their own thread. This is less confusing and ensures that you get prompt help.

I see you have already gotten help with the problem, so I will leave you in the expert hands of SuperSonic4. :D

 

[HallsofIvy]

Another way: starting from the equations
18E+ 17V= 306.50
19E+ 15V= 298.50

Multiply the first equation by 15 (the coefficient of V in the second equation) and multiply the second equation by 17 (the coefficient of V in the first equation) to get
270E+ 255V= 4597.5
323E+ 255V= 5074.5

Now, the two equations have the same coefficient for V and subtracting the first equation from the second eliminates V: 53E= 477 so that E= 477/53. Now put that back into either of the original equations to get a single equation for V.

 

[CellCoree]

To find the cost per pound of the espresso and vanilla flavored coffee beans, we can set up a system of linear equations.

Let E be the cost per pound of espresso flavored coffee beans and V be the cost per pound of vanilla flavored coffee beans.

From the given information, we can write the following equations:

18E + 17V = 306.50
19E + 15V = 298.50

To solve this system, we can use the method of substitution. From the first equation, we can rearrange it to get:

E = (306.50 - 17V)/18

We can then substitute this value of E into the second equation:

19[(306.50 - 17V)/18] + 15V = 298.50

Simplifying this equation, we get:

306.50 - 17V + 15V = 298.50
-2V = -8
V = 4

Now, we can substitute this value of V into the first equation to find the value of E:

18E + 17(4) = 306.50
18E = 306.50 - 68
18E = 238.50
E = 13.25

Therefore, the cost per pound of espresso flavored coffee beans is \$13.25 and the cost per pound of vanilla flavored coffee beans is \$4. We can double check this by substituting these values into the original equations:

18(13.25) + 17(4) = 306.50
19(13.25) + 15(4) = 298.50

Both equations hold true, so our solution is correct. The cost per pound of espresso and vanilla flavored coffee beans is \$13.25 and \$4, respectively.