*System of equations AND Inequation* (Need help)

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In summary, Alex hikes up a mountain at a speed of 12km/hour and returns home at a speed of 18km/hour. The difference in time is 15 minutes.
  • #1
thinkies
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[SOLVED] *System of equations AND Inequation* (Need help)...

Homework Statement



Alex is a professionel athlete. One day, he leaves for mountain hiking and ascends a coast to an average speed of 12 km/hour. On his way back, he borrows the same coast and descends to an average speed of 18 km/hour. The descent lasted 15 minutes less than the ascension.

Find the coast length.

P.S.: *Sorry for the typo, I had to translate this problem from french to english =.=...

Homework Equations



...

The Attempt at a Solution



Where do I even start with this...
This problem may be easy for some of you...well its a typical 9th grade question...

(P.S.: I did attempt something..however I am not sure.Waiting for your answers/tips/advices...)
 
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  • #2
umm, this problem comes from one of my math book chapter: System of Equations and Inequation
 
  • #3
What's a 'coast'? No matter. Call it's length L. Then the first time is L/(12km/hr), and the second is L/(18km/hr). The difference of those two times is 15 min. Write an equation expressing that and solve for L.
 
  • #4
Since the speed is expressed in Km/hour...and if i convert this in km/minute...it would become:

12/60=0.2km/minute

18/60=0.3km/minute

So the equation is...

0.2m = 0.3m - 15?

I solve this and find m...it still sounds weird...o.0
 
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  • #5
how would the above formula(the 1 i just posted) can help find the length...

is it normal that I am facing(being a 9th grade student)difficulties in this problem,is it suppose to be an easy-level problem...?
 
  • #6
also,if i solve the above equation,it gives 150! A guy siriously can not be hiking on a 150 KM track! o.0...
 
  • #7
bump.....
 
  • #8
so,what do you think dick?
 
  • #9
Ah, 'coast'='slope' or 'hill'. The equation becomes L/(0.2km/min)-L/(0.3km/min)=15min. The only unknown is L. Solve for it.
 
  • #10
thinkies said:
12/60=0.2km/minute

18/60=0.3km/minute

So the equation is...

0.2m = 0.3m - 15?

I solve this and find m...it still sounds weird...o.0
It would help if you expressed units all the way through. If you had done this you would see that the equation is incorrect because the units don't match up. The '0.2' is more precisely stated as '0.2 km/minute', 'm' as 'm km', and 15 as '15 minutes'. (This makes your 'm' unitless -- just a number.) Being explicit,

m*0.2 km^2/minute = m*0.3 km^2/minute - 15 minutes

and this doesn't make a bit of sense, which is why your answer doesn't make a bit of sense. Re-read Dick's instructions.
 
  • #11
you equationd doesn't work.

"The descent lasted 15 minutes less than the ascension"

we don't even know how many minutes he took for the ascension...

and we are looking for the length...ah i don't see where you're really getting...o.0
 
  • #12
D H said:
It would help if you expressed units all the way through. If you had done this you would see that the equation is incorrect because the units don't match up. The '0.2' is more precisely stated as '0.2 km/minute', 'm' as 'm km', and 15 as '15 minutes'. (This makes your 'm' unitless -- just a number.) Being explicit,

m*0.2 km^2/minute = m*0.3 km^2/minute - 15 minutes

and this doesn't make a bit of sense, which is why your answer doesn't make a bit of sense. Re-read Dick's instructions.

um...0.2 kilometres/min doesn't seem wrong at all...in fact,its ok if i leave the unit as a KM rather transforming it into m...
 
  • #13
You don't know the time for the ascent. But if the length is L, then that time can be written L/(0.2km/min). Time=distance/velocity.
 
  • #14
sorry if my reasonning seems mindless and newbish...=.=
 
  • #15
thinkies said:
sorry if my reasonning seems mindless and newbish...=.=

There's nothing wrong with your reasoning so far. You just haven't finished. Read the advice you are getting carefully and think about it before posting again.
 
  • #16
so...

L/(0.2km/m) - L(0.3km/m) = 15

L/-0.1 = 15

L = -1.5

I messed this up..for sure
 
  • #17
Sure did. Nothing wrong with the concept. Just bad algebra. L/0.2-L/0.3 is not L/(-0.1). You are subtracting fractions, you need to find a common denominator.
 
  • #18
Dick said:
Sure did. Nothing wrong with the concept. Just bad algebra. L/0.2-L/0.3 is not L/(-0.1). You are subtracting fractions, you need to find a common denominator.

HAHA,Sooorrrry it just came up in my mind,I rushed up with this...must pay attention next time.Thx

Now...

L/(0.2Km/m) - L/(0.3km/m) = 15

L/(0.6 km/m) - L/(0.6km/m) = 15

0L = 9 KM?
 
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  • #19
You really need some practice. L/(0.2)=(3/3)*L/(0.2)=3L/(0.6). Multiply the second one by 2/2.
 
  • #20
Dick said:
You really need some practice. L/(0.2)=(3/3)*L/(0.2)=3L/(0.6). Multiply the second one by 2/2.

L = 9km ?

yup,i rushed up witht he above,forgot to multiply the L's...
 
  • #21
thinkies said:
L = 9km ?

yup,i rushed up witht he above,forgot to multiply the L's...

? 3L/0.6-2L/0.6=L/0.6=15. What's L?
 
  • #22
Yes. L=9km. Practice a lot more examples.
 
  • #23
Dick said:
? 3L/0.6-2L/0.6=L/0.6=15. What's L?

L/0.6 = 15

L = 15 X 0.6

L = 9km

L= Lenght of the slope...
 
  • #24
actualy I am pretty gud with algebra..i simply 'rushed' things ...
 
  • #25
thinkies said:
actualy I am pretty gud with algebra..i simply 'rushed' things ...

Ok. Then work on your english spelling. :)
 
  • #26
you meantioned earlier that time=distance/velocity...

didnt we just calculated time...instead of lenght?..
 
  • #27
if i remember well...the official answer that was written in the correction was something else then 9 km :(...
 
  • #28
thinkies said:
you meantioned earlier that time=distance/velocity...

didnt we just calculated time...instead of lenght?..

We solved the equation for the difference of times for the unknown distance. Didn't we? You are writing a lot faster than you are thinking.
 
  • #29
But,my questions was what is the lengh of the slope...and this is how I've been guided throughout(by some of you,please don't take it offensively)..=.=...
 
  • #30
The length of the slope is L. You correctly computed L=9km. I don't know what you are talking about.
 
  • #31
well..thx a lot for the help and sorry for the confusion.I guess i simply didnt translate the problem well (from french to english)...thx though
 

1. What is a system of equations and inequations?

A system of equations and inequations is a set of two or more equations and/or inequalities that are related to each other and have a common set of solutions. These equations and inequalities can involve multiple variables and can be solved simultaneously to find the values of the variables that satisfy all of the equations and inequalities in the system.

2. How do you solve a system of equations and inequations?

To solve a system of equations and inequations, you can use various methods such as substitution, elimination, or graphing. The goal is to find the values of the variables that satisfy all of the equations and inequalities in the system. Once you have found these values, you can check if they satisfy all of the equations and inequalities in the system.

3. What is the difference between equations and inequations?

Equations are mathematical statements that show that two expressions are equal, while inequations show that two expressions are not equal. Inequations also include symbols such as <, >, ≤, and ≥ to indicate the relationship between the two expressions. Equations and inequations can both be used to represent real-world situations and solve problems.

4. Can a system of equations and inequations have no solution?

Yes, a system of equations and inequations can have no solution. This means that there is no set of values for the variables that satisfy all of the equations and inequalities in the system. Graphically, this would result in parallel lines or no intersection points on a graph. In this case, the system is said to be inconsistent.

5. How are systems of equations and inequations used in real life?

Systems of equations and inequations are used in many real-life situations, such as in business, economics, and science. They can be used to model and solve problems involving multiple variables, such as finding the optimal solution for a business or determining the best combination of ingredients for a recipe. They can also be used to analyze and make predictions about various systems or processes.

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