Question

at a premier pet store the cost of three cans of dog food and two raw hide bones is 8.60 if a customer buys four cans of dog food but returns one raw hide bone the cost is 7.25 what is the cost of each can of dog food and each raw hide bone?

Answer 1

To solve for the cost of one can of dog food and one raw hide bone, we set up a system of equations and determined that each can of dog food costs $2.10, and each raw hide bone costs $1.15.

Step-by-step explanation:

To solve this problem, we can set up two equations based on the given information. Let c represent the cost of one can of dog food, and b represent the cost of one raw hide bone.

The first equation comes from the first scenario where three cans of dog food and two raw hide bones cost $8.60:

3c + 2b = 8.60

The second equation comes from the second scenario where four cans of dog food are bought and one raw hide bone is returned, resulting in a cost of $7.25:

4c - b = 7.25

To solve the system of equations, we can multiply the second equation by 2 to eliminate b when added to the first equation:

8c - 2b = 14.50

Adding this equation to the first equation provides us:

3c + 2b = 8.60
+
8c - 2b = 14.50
_____________
11c = 23.10

Dividing both sides by 11 gives us the cost of one can of dog food:

c = 2.10

Now, we can solve for b by substituting c into one of the original equations:

3(2.10) + 2b = 8.60
6.30 + 2b = 8.60
2b = 8.60 - 6.30
2b = 2.30

Dividing both sides by 2 gives us the cost of one raw hide bone:

b = 1.15

Thus, each can of dog food costs $2.10, and each raw hide bone costs $1.15.

Answer 2

The cost of three cans of dog food and two raw hide bones is, 6.80

If the customer buys four cans of dog food but returns one raw hide bone the cost is 7.25.

So, Let x be the cost of one can of dog food and y be the cost of one raw hide bone. Then we have,

`\begin{gathered} 3x+2y=8.60\ldots\text{.}(1) \\ 4x=7.25\ldots\text{.}(2) \end{gathered}`

From the equation 2 we get,

`x=(7.25)/(4)=1.8125\approx2`

Substituting the value of x in equation 1, we get,

`\begin{gathered} 3*(7.25)/(4)+3y=8.60 \\ 3*1.8125+3y=8.60 \\ 3y=8.6-5.4375=3.1625 \\ y=(3.1625)/(3)=1.05\approx1 \\ \end{gathered}`