Respuesta :
First, line up your facts between the two companies:
Company A: $110 + $.20 per mile
Company B: $50 + $.60 per mile
In this situation, we need to use a basic algebra equation. Notice that each company has a daily flat rate ( Company A: $100, Company B: $50 ). That flat rate is the first part of each side of your equation. Since we are going based on only one day, we don't have to include a variable attached to that daily rate. Now, each company has a different rate per mile driven. Company A's rate is 20 cents per mile, and B's is 60 cents per mile. Since each cent amount (20 and 60) goes based on the amount of miles driven, we can attach a variable as "m" - meaning "miles driven". Set up your First part of this problem like this:
Company A: | Company B:
$110 + $.20 m (per mile)|$50 + $.60 m (per mile)
Now, each of these companies represents one side of the total equation. Look at the first photo. Set up your new equation like that, since each company represents one side of the equation.
Now proceed like a normal algebra equation.
110 + .20 m = 50 + .60 m
-50 -50
60. 0
60 + .20 m = .60 m
Since we subtracted 50 from each side of the equation, we are one step closer to isolating the variable: m.
60 + .20 m = .60 m
-.20 m -.20 m
60 + 0. = .40 m
60 = .40 m
Now that you have the variable and it's coefficient completely isolated, we can now divide the 60 from the left side by .40, which will ultimately leave us with the value of "m".
60 ÷ .40 = 150.
150 = miles
So, at 150 miles that have been driven, the cost that it would be to do each Company's service is 150 miles. At 150 miles, the cost for each Company is the same. Be sure to double check your answers!
Company A: $110 + $.20 per mile
Company B: $50 + $.60 per mile
In this situation, we need to use a basic algebra equation. Notice that each company has a daily flat rate ( Company A: $100, Company B: $50 ). That flat rate is the first part of each side of your equation. Since we are going based on only one day, we don't have to include a variable attached to that daily rate. Now, each company has a different rate per mile driven. Company A's rate is 20 cents per mile, and B's is 60 cents per mile. Since each cent amount (20 and 60) goes based on the amount of miles driven, we can attach a variable as "m" - meaning "miles driven". Set up your First part of this problem like this:
Company A: | Company B:
$110 + $.20 m (per mile)|$50 + $.60 m (per mile)
Now, each of these companies represents one side of the total equation. Look at the first photo. Set up your new equation like that, since each company represents one side of the equation.
Now proceed like a normal algebra equation.
110 + .20 m = 50 + .60 m
-50 -50
60. 0
60 + .20 m = .60 m
Since we subtracted 50 from each side of the equation, we are one step closer to isolating the variable: m.
60 + .20 m = .60 m
-.20 m -.20 m
60 + 0. = .40 m
60 = .40 m
Now that you have the variable and it's coefficient completely isolated, we can now divide the 60 from the left side by .40, which will ultimately leave us with the value of "m".
60 ÷ .40 = 150.
150 = miles
So, at 150 miles that have been driven, the cost that it would be to do each Company's service is 150 miles. At 150 miles, the cost for each Company is the same. Be sure to double check your answers!