# Video: US-SAT05S4-Q23-942197061237

The given table shows information on ticket prices, in dollars, and distances, in meters, of flights from Detroit to different destinations, as seen on the website of an airline company. The linear relationship between the price of a ticket 𝑝 in dollars and the distance of the flight 𝑑 in thousands of meters can be modeled by a linear function 𝑝(𝑑). Which of the following functions represents the relationship? [A] 𝑝(𝑑) = 1.5𝑑 − 30 [B] 𝑝(𝑑) = 2𝑑 − 820 [C] 𝑝(𝑑) = 0.6𝑑 + 152 [D] 𝑝(𝑑) = 0.2𝑑 + 140

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### Video Transcript

The given table shows information on ticket prices, in dollars, and distances, in meters, of flights from Detroit to different destinations, as seen on the website of an airline company. The linear relationship between the price of a ticket 𝑝 in dollars and the distance of the flight 𝑑 in thousands of meters can be modeled by a linear function 𝑝 of 𝑑. Which of the following functions represents the relationship? Is it A) 𝑝 of 𝑑 is equal to 1.5𝑑 minus 30. B) 𝑝 of 𝑑 is equal to two 𝑑 minus 820. C) 𝑝 of 𝑑 is equal to 0.6𝑑 plus 152. Or D) 𝑝 of 𝑑 is equal to 0.2𝑑 plus 140.

We’re told in the question that the relationship is linear. This means that the equation must be of the form 𝑦 equals 𝑚𝑥 plus 𝑏. Where 𝑚 is the slope or gradient and 𝑏 is the 𝑦-intercept. It is important to note that 𝑑 is measured in thousands of meters. This means that the value for 𝑑 for each city is 260, 820, 720, 490, and 380. The ticket price in dollars is our 𝑝 of 𝑑 or 𝑝-value. We could pick any two cities to substitute into the equation. In this case, we’ll pick Columbus and New York. We can substitute these values into the equation 𝑦 equals 𝑚𝑥 plus 𝑏. Where 𝑑 is the 𝑥-value and 𝑝 of 𝑑 is the 𝑦-value.

Substituting in our values for Columbus gives us 192 is equal to 260𝑚 plus 𝑏. We will call this equation one. For New York, our equation becomes 304 is equal to 820𝑚 plus 𝑏. We will call this equation two. We can solve this pair of simultaneous equations to calculate the value of 𝑚 and 𝑏 by elimination. We can subtract equation one from equation two. 304 minus 192 is equal to 112. 820 minus 260 is equal to 560. Therefore, 820𝑚 minus 260𝑚 is equal to 560𝑚. 𝑏 minus 𝑏 is equal to zero. So the 𝑏s cancel. We can now divide both sides of this equation by 560. 112 divided by 560 simplifies to one-fifth. This is equal to 0.2. The right-hand side simplifies to 𝑚 as the 560s cancel. Therefore, 𝑚 is equal to 0.2.

We can immediately see that the only equation with a slope or gradient of 0.2 is option D. Therefore, this must be the correct answer. However, it is worth substituting this number back in to calculate 𝑏. So we can check our answer. We need to substitute our value for 𝑚 back into one of the equations. In this case, we’ll substitute into equation one. This gives us 192 is equal to 260 multiplied by 0.2 plus 𝑏. 260 multiplied by 0.2 is equal to 52. Subtracting 52 from both sides of this equation gives us a value of 𝑏 equal to 140. We can therefore say that the linear relationship in the form 𝑦 equals 𝑚𝑥 plus 𝑏 is 𝑦 equals 0.2𝑥 plus 140. Replacing the variables used in this question, 𝑝 of 𝑑 is equal to 0.2𝑑 plus 140. The price in dollars is equal to 0.2 multiplied by the distance in thousands of meters plus 140.

We could check this answer by substituting in the values for Philadelphia, Louisville, or Indianapolis. The cost of a ticket to Philadelphia was 284 dollars. And the distance was 720000 meters. Substituting the correct values into the formula gives us 284 is equal to 0.2 multiplied by 720 plus 140. This calculation is correct. And we could also check as mentioned for Louisville and Indianapolis.

Looking back at our four options, there is a quick way we could’ve eliminated option A and option B. Both of these had a negative 𝑦-intercept. This would mean that when the distance was small or zero, the price or cost would be negative, which is impossible. Therefore, these two options are definitely incorrect. The correct answer of 0.2𝑑 plus 140 tells us that there is a basic cost of 140 dollars no matter what the distance. And that we’re charged 20 cents or 0.2 dollars for every 1000 meters in distance.