Problem with simultaneous equation

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dontknow

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Aug 5, 2014
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Hello, I'm having trouble with what I believe is a simultaneous equation. I have referred to books but can't quite understand how to solve it.
The problem is this:

"A tank is being filled from a tap at 125L/per hour. If it were filled at 150L/per hour the tank would fill in 1 hour and 20 minutes sooner.
What is the volume of the tank?"

I've made these steps so far:

Volume= time x L per hour
Let y = volume
Let x = time

eqn1: y=125x
eqn2: y=150(x-3/4)

What would be the logic behind solving this and what steps should I take?
Is substitution, elimination or tabling useful?

Any help would be great.
 
dontknow said:
"A tank is being filled from a tap at 125L/per hour. If it were filled at 150L/per hour the tank would fill in 1 hour and 20 minutes sooner. What is the volume of the tank?"

I've made these steps so far:

Volume= time x L per hour
Let y = volume
Let x = time

eqn1: y=125x
eqn2: y=150(x-3/4)
Click to expand...
What is the logic for "x - 3/4"? What does this mean? If "x" stands for "the total time needed to fill the tank at the slower rate, stated in hours", then "x - 3/4" is "forty-five minutes less than that". Is this what you meant? If so, why?

dontknow said:
What would be the logic behind solving this and what steps should I take?
Is substitution, elimination or tabling useful?
Click to expand...
I don't know what "tabling" is in this context but, yes, any method for solving systems of linear equations would be helpful for solving this system of linear equations. ;)
 
dontknow said:
Hello, I'm having trouble with what I believe is a simultaneous equation. I have referred to books but can't quite understand how to solve it.
The problem is this:

"A tank is being filled from a tap at 125L/per hour. If it were filled at 150L/per hour the tank would fill in 1 hour and 20 minutes sooner.
What is the volume of the tank?"

I've made these steps so far:

Volume= time x L per hour
Let y = volume
Let x = time

eqn1: y=125x
eqn2: y=150(x-3/4)

What would be the logic behind solving this and what steps should I take?
Is substitution, elimination or tabling useful?

Any help would be great.
Click to expand...
The simplest logic would be to say that since y is equal to both 125x and 150(x- 3/4) (I am simply copying your equations. I am not saying if they are right or wrong) then they are equal to each other: 125x= 150(x- 3/4). Another way would be to say that is a= b and c= d then a- c= b- d. So that y- y= 0= 125x- 150(x- 3/4).
 
It was a typo, 3/4 should have been 4/3 which would represent the 1h 20m. Tabling seems to not be useful so don't worry about it.
After spending way too long on such a simple problem I've solved it. If you're curious I can explain the solution to you. :wink:

Is it possible to delete this post? If not, admin feel free to do so.
 
Simultaneous equations

It is interesting to consider the essence of a pair of equations being "simultaneous".

When are two equations related by being "simultaneous"?

How can one create a pair of simultaneous equations?

Is it a matter of operating on two equations using, say, addition or multiplication, composition? Is another relation needed. What does "substitution" have to do with it?

If one knew how to create a simultaneous equation then it might make more clear the essence of a physical situation (a word problem) when they are needed and what to build them around as well as the fundamental method that underlies procedures for solving them.

I don't believe I have ever had a math class that asked me to just make up a pair of simultaneous equations describing any something, at least not explicitly so that I would realize what I was doing. I sort of wish they had.
 
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