Every time dad fills his truck he reset his millage counter, from the gas station to home it is 2.3 KM. His counter reads 116.8 KM; he has only made four trips, all to work, however the fourth trip took longer than the other three. So my resulting formula is:

114.5 - 3X = Y; Y > X

I ended up with:

Y X3 1 - ----- = ----- 114.5 114.5

Which is pretty much worthless, all I need is a ratio of Y:X.

]]>Every time dad fills his truck he reset his millage counter, from the gas station to home it is 2.3 KM. His counter reads 116.8 KM; he has only made four trips, all to work, however the fourth trip took longer than the other three. So my resulting formula is:

114.5 - 3X = Y; Y > X

I ended up with:

Y X3 1 - ----- = ----- 114.5 114.5

Which is pretty much worthless, all I need is a ratio of Y:X.

]]> Um.. what exactly is it that you're trying to find out??

So what are your trying to figure out? The mileage on his truck? The distance to work? The second distance to work?

But when does his counter read 116? At home? at work? Juuuust before filling up?

The trips to work -- roundtrips or single?

You're going to have to be a little more specific.

I think I figured it out. The 116.8 is the counter at home, after four trips to work and back, three of which were the same length (x), one of which was a detour (y). The counter includes the distance from the gas station to home before the first trip, which we subtract to get a total distance of 114.5 for the four trips together:

3x + y = 114.5

We also know:

y > x > 0

The desired result is the ratio between the different trip lengths. Unfortunately, I don't think it's possible to get an answer; there is not sufficient information. My reasoning: you can reform the first equation to:

y = 114.5 - 3x

And now you can insert different guesses for x to get different corresponding values for y that satisfy the equations by result in different rations between x and y:

x = 20 -> y = 54.5 -> y/x = 2.725

x = 25 -> y = 39.5 -> y/x = 1.58

Using the inequation y > x > 0, the best you could do is get an upper and lower bound for the ratio.

]]> This feature is not included in Math Basic. You need to upgrade to Math Professional to solve that problem.

Using the inequation y > x > 0, the best you could do is get an upper and lower bound for the ratio.

This is a function : 3x + y = a constant. Try drawing a graph... ]]>

Wait until he gets back from work tomorrow, check the meter again. Substract 114.5 and now you know what x is!

- Al

]]>I suspect that it was in fact your mum who put up the problem - just as she put up the travelling salesman problem last week concerning gin joints & bordellos in the area.

Expect an alimony-based problem some time soon.

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@Hatshepsut said:

I suspect that it was in fact your mum who put up the problem - just as she put up the travelling salesman problem last week concerning gin joints & bordellos in the area.

Expect an alimony-based problem some time soon.

LOL!

ANd brazzy's solution is correct. Any problem with 2 unknowns is going to be "unsolvable" in the strictest sense. you can solve for one and have a range for the other one (IE you know what it CAN'T be.)

Texas Instruments designs calculators that do this sort of math I think they're called polynomials or something. It's been well over 10 years since i last took a math class, so don't quote me!

]]>this is a simple linear equation. It will result in a line that represents all possible ratios. use y > x > 0 to narrow the field to a line segment. That is all you can do.

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