Answer:
-5x + y = -1 or y = 5x - 1
Step-by-step explanation:
5x - y = 9
-5x - 5x
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-y = -5x + 9
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-1 -1
y = 5x - 9
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-6 = 5[-1] + b
-5
-1 = b
y = 5x - 1
If you want it in <em>Standard </em><em>Form</em>:
y = 5x - 1
-5x -5x
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-5x + y = -1 >> Line in <em>Standard</em><em> </em><em>Form</em>
I am joyous to assist you anytime.
-9.5
I think I might be wrong
gives an average cost per unit, if we want to produce x of them.
So for example, we want to produce 500 toy cars for our store, and we need a price per unit (per 1 toy car). What we do is we calculate C(500).
So to calculate the cost of one unit when producing 1250, we calculate C(1250)

$ is the cost of 1 toy car.
Answer
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
Now, the system of equations in matrix form will be:
![\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%26%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26%5C%5Cy%26%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%26%5C%5C7%26%5Cend%7Barray%7D%5Cright%5D%20)
Next, we are going to find the determinant:
![D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}](https://tex.z-dn.net/?f=D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%281%29%28%20%5Cfrac%7B1%7D%7B25%7D%29%20-%20%281%29%28%20%5Cfrac%7B1%7D%7B65%7D%20%29%3D%20%5Cfrac%7B8%7D%7B325%7D%20)
Next, we are going to find the determinant of x:
![D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8](https://tex.z-dn.net/?f=%20D_%7Bx%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%261%5C%5C7%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%28375%29%28%20%5Cfrac%7B1%7D%7B25%7D%20%29-%281%29%287%29%3D8)
Now, we can find x:

Now that we know the value of x, we can find y:

Remember that time equals distance over velocity; therefore, the time on the highway will be:

An the time on the city will be:

We can conclude that the bus was five hours on the highway and two hours in the city.