# Word problem equation linear growth and slope

1. May 24, 2016

### late347

1. The problem statement, all variables and given/known data
truck moves oils (lubricating oil presumably) to gas stations.
At full load the truck transportation cost is 300$when the travelling distance is 120km and the transportation cost is 350$ when the travelling distance is 165km
At this interval, the costs change linearly. You may use calculater for aid.

a) calculate the cost when the distance travelled is 150km

b) assuming the cost and distance change linearly even beyond the original interval, calculate the new distance, for when the cost would be 1000$c) when do the costs fall, below 2$ per kilometer

2. Relevant equations
k = deltay/deltax

3. The attempt at a solution

equation for straight line is of the form
Also it should be noted that the costs do actually grow, when the distance travelled actually grows... there was something fishy about the ratio though. I think we must probably find the correct constant value for the b term in the equation

y= k * x + b

k= (350-300) / (165- 120)
k= 10/9

it is said that the interval has linear growths, I think we should probe the starting value, to see what the constant b would have been

300 = (10/9) * 120 + b
b= 500/3

y=( 10/3 ) * x + (500/3)

probe for the end value in interval (assuming end point and beginnning point are within the interval)

we achieve y= 350
When the values x= 165 and k= 10/9 and b = 500/3 is used

a) cost = 333,33 , distance travelled = 150km

b) cost = 1000$distance traveleld = 750km c) I need help for this part I think My teacher sent me the different answers as mine. She sent me the answers to these. Teacher had calculated k = (350-300) / (165-120) = 9/10 Correspondingly teacher had different value for the b term. I don't think you end up with 9/10 = 50/45 in any mathematical fashion right there... but I don't know. I think earlier my math teacher said that brackets are calculated first. (350-300) / (165-120) = 50/45 50/45 fraction can be shrunken without changing its value ( I don't know what the procedure is called in English language, but fractions can shown in different forms, without changing its value, when you multiply or divide both the numeratorr and denominator by the same number which would not be 0) 10/ 9 = 50/45 because 50/5= 10 45/ 5 = 9 The question clearly stated that the costs change linearly with the distance travellled. Cost grows, and distance grows. I don't think it is possible that the cost and the distance are inversely correlated between each other. I I'm pretty confused about whether or not my teacher was correct or whether I was correct. I could use help to the c part of the problem. How do I make sure that I find when the cost per km, drops to below (2$ / km)?

2. May 24, 2016

### SteamKing

Staff Emeritus
There is a typo in the equation above. Can you spot it?
The answers to a) and b) look good.
Your equation is correct for finding the cost of the transport by using the distance traveled as the independent variable.

Your teacher seems to be using the cost of the trip as the independent variable, and then presumably finding the distance from that cost.
Of course you do. Does 50 have 9 as an even factor? Does 45 have 10 as an even factor?

If the answer to either question is a big fat No, then 9/10 ≠ 50/45

BTW, 9/10 < 1 and 50/45 > 1, so there is no way that 9/10 = 50/45
This is called simplifying or reducing a fraction to the lowest terms.
5 is a common factor in both the numerator and the denominator of the fraction, and common factors can be canceled out to reduce the fraction to the lowest terms.
If you take the distance traveled in the first two examples and divide that figure into the cost of the trip, you will find out how much you are spending to transport the oil on an average dollars per kilometer basis.

For example, cost = $300 and distance = 120 km, therefore the cost to distance ratio is$300/120 km = $2.50 / km What question c) wants to know is, using your cost and distance formula, how long a trip would the truck need to take so that when the total cost is divided by the distance, that ratio, Cost / Distance, is not greater than$2/km.

3. May 24, 2016

### BvU

Hello

As a word of advice: making a drawing at such a point can be very illuminating !

4. May 24, 2016

### late347

my teacher got such equation for the line
cost = 0,9 * distance + 192

cost1= 327dollars at 150km
distance2 = 898km at 1000dollar

--------------------------------

My own results were

Y= (10/9) x + 500/3

It can be shown in better way for addition purposes as follows
y= (10/9) X + 1500/9

333,33 dollar at 150km
and 1000 dollar at 750km

I think the necessary ratio for c) part is

2 dollars / 1 kilometer = 2/1

I gleaned at my teacher's notes and it seems she did it such that

y/x = 2/1

multiply crossing in crossing fashion, (don't know what this is called in English mathematics properly)

2x = y

2x= (10/9) X + 1500/9

18x =10x+ 1500

8x= 1500
x= 187,5

5. May 24, 2016

### SteamKing

Staff Emeritus
You got the correct equation to figure transport cost given distance, and your math teacher's mistake is partially hidden by the fact that the slope of his equation is close to yours.
You have used the correct procedure for finding the slope, since it gives a value in dollars / km, while the slope for your teacher's equation is actually km / dollar.

As BvU suggested earlier, a picture is worth a lot of calculations.
It's called cross-multiplying