Answer:
![y=0.2x+284](https://tex.z-dn.net/?f=y%3D0.2x%2B284)
Step-by-step explanation:
Let x be the distance driven, d-distance and C our constant.
Our information can be presented as:
![y=mx+c\\\\450=830x+C\ \ \ \ \ \ \ \ \ eqtn 1\\\\380=480x+C\ \ \ \ \ \ \ \ \ eqtn 2](https://tex.z-dn.net/?f=y%3Dmx%2Bc%5C%5C%5C%5C450%3D830x%2BC%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20eqtn%201%5C%5C%5C%5C380%3D480x%2BC%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20eqtn%202)
#Subtracting equation 2 from 1:
![70=350x\\x=0.2](https://tex.z-dn.net/?f=70%3D350x%5C%5Cx%3D0.2)
Hence the fixed cost per mile driven,
is $0.20
To find the constant,
we substitute
in any of the equations:
![450=830x+C\ \ \ \ \ \ \ X=0.2\\\therefore C=450-830\times0.2\\=284](https://tex.z-dn.net/?f=450%3D830x%2BC%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20X%3D0.2%5C%5C%5Ctherefore%20C%3D450-830%5Ctimes0.2%5C%5C%3D284)
Now, substituting our values in the linear equation:
#y=cost of driving, x=distance driven
Hence the linear equation for the cost of driving is y+0.2x+284