An analogy: For this set of questions, choose between two cars that we will call "A" and "B." Car A is a two-door coupe. It gets 60 miles per gallon and has a 10-gallon tank. Car B is an SUV. It gets 2 miles per gallon and has a 60-gallon tank. Suppose both cars embark on a very long journey together, side by side, each with a full tank of gas. Which car holds more fuel?

Car B, the SUV, obviously holds more fuel after 0 driven miles, 60 gallons, while car A holds 10 gallons after 0 miles.

let's speak about miles with the abbreviation x.

sk at x=0 Car B got more fuel.

Car A gets 60 Miles per gallon, it's consuming 1/60 gallon per mile. The amount of fuel decreases by 1/60 per mile

After each mile, or each x-step, Car A has

This can be expressed mathematically:

A(x) = -1/60*x + 10

The -1/60x is, again, the consumption per mile, the +10 is the full tank at the start.

A(0)= -1/60*0 + 10

=10

For Car B the consumption is 1/2 gallons per mile and the tank holds 60 gallons of fuel, wich ammount is decreases with traveled distance.

B(x) = -1/2*x + 60

See that we could draw these two functions as lines in a coordinate system. (I did it with desmos btw). This step is just to help understand.

After some distance/ amount x of miles, Car A will hold more fuel than car B, because it drives more efficiently.

In the picture we can see that this is after about 100 miles and we could guess the amount of fuel, but I won't. Let's calculate it and then answer the question precisely.

We get the intersection of the two lines by setting them equal, a(x)=b(x)

-1/60x + 10 = -1/2x + 60

Exactly one point lies, equally, on both lines

So let's solve for x

Further on the right, each step is explained. we have to do the same operation on both sides of the equation, just to not break it.

-1/60x + 10 = -1/2x + 60 | -10

-1/60x = -1/2x + 50 | + 1/2x

-1/60x + 1/2x = 50 | *60

-1x + 30x = 3000

29x = 3000 | /29

x = 103.448275862

so after about 103.45 miles, Car B holds less fuel than Car A (and Car A more than Car B), until both are empty. Car B could therefore drive more miles

We can also ask how much fuel is left in both tanks at the intersection point by plugging the x-value we got into one of the equations (doesn't matter which one, since the point is the same)

a(103.4482) = -1/60 * 103.4482 + 10 = 8.28

If you want it to be precise, stay with fractions

a(3000/29) = -1/60 * 3000/29 + 10

= -50/29 +10

= 9 -21/29

= 8 + 8/29

Finding out how far each car could travel would be achieved by setting their fuel-equation equal to 0, because that's the value of it when they run of of fuel. For example 0 = -1/60x+10, then you (just) solve for x.

I did overshoot the given problem to offer offer you understanding that's spans more of the topic you are dealing with

(would really appreciate the brainliest)

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