A nut mixture of peanuts and pecans at a small fair is $1.00 per pound of peanuts and $2.85 per pound of pecans. Over the entire day, 97 pounds of the nut mixture were sold for $254.25. If p is the number peanuts and n is the number of pecans, then the system of equations that models this scenario is:

p+n=97
p+2.85n=254.25

Determine the correct description and amount of pounds for peanuts and pecans that were sold.
A (12.0, 85.0)There were 12.0 pounds of peanuts and 85.0 pounds of pecans sold at the fair.
B (75.0, 22.0)There were 75.0 pounds of peanuts and 22.0 pounds of pecans sold at the fair.
C (85.0, 12.0)There were 85.0 pounds of peanuts and 12.0 pounds of pecans sold at the fair.
D (22.0, 75.0)There were 22.0 pounds of peanuts and 75.0 pounds of pecans sold at the fair.

I NEED HELP ASAP

A - 12.0, 85.0
You can solve this system of equations by subtracting p+2.85n-p-n=254.25-97 and if you solve that, you get 1.85n=57.25 and that can be simplified to n=85

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slicergiza slicergiza · Answer 2 · 2019-10-21T09:09:03+02:00

Answer:

A. (12.0, 85.0)There were 12.0 pounds of peanuts and 85.0 pounds of pecans sold at the fair.

Step-by-step explanation:

Here, the given system of equations,

p + n = 97 ------(1)

p + 2.85n = 254.25 -----(2)

Where,

p = number of pounds of peanuts,

n = number of pounds of pecans,

Equation (2) - equation (1),

We get,

1.85n = 157.25,

[tex]\implies n = \frac{157.25}{1.85}=85[/tex]

From equation (1),

p + 85 = 97 ⇒ p = 97 - 85 = 12,

Thus, there were 12.0 pounds of peanuts and 85.0 pounds of pecans sold at the fair.

i.e. the solution of the above system of equation is (12.0, 85.0)

OPTION A is correct.