Problem with simultaneous equations

[Marcus Clayson]

 

[Deleted member 4993]

View attachment 33151

Why do you say your solution "is obviously wrong"?

 

[Dr.Peterson]

Please show us the problem, so we can tell why you think the answer is wrong!

 

[Steven G]

No one here has any idea where you got your equations from. The work you have shown is correct but not complete. Now solve for y.

Twice you said you don't think that you set up the equations correctly. That might be true, so don't you think that it might be helpful if you show the whole problm so we can tell you why your equations are not correct, if in fact they are.

 

[Harry_the_cat]

Without seeing the actual problem, it is very hard to see where you have gone wrong, if in fact you have. My guess is is that you are using metres on on side of the equation and kilometres on the other. Check that your units match up.

 

[Marcus Clayson]

First of all, can I say thank you to those who replied to my post so quickly.
Secondly, I have to admit that, true to form, my laptop has sent the wrong files - or rather, one page twice instead of pages 1 AND 2. I am attaching pages 1 AND 2 now so that people can hopefully see the original task. ...865 is page 1 and ...866 page 2. I don't know how to make the attachments big enough to read before sending, so I am banking on the original file numbers being correct. I'm not a fan of computers

 

[Steven G]

1)A motorist travels at x km at 40km/hr and y km at 50km/hr in 2 1/2hrs.

2)Also 6x at 30km/hr and 4ykm at 50km/hr in 2 1/2 hrs in 14 hrs.

Find x and y

We have the time in each case, so lets solve for time.

We know that rate*time = distance or r*t = d. So t = d/r

So equation 1equals 2 1/2. It also equals d/r. You have both d and r (in terms of x and y). Can you write the equation that corresponds to 1? How about to 2?

The main reason your equations are wrong is because t ≠ r*d !!! You even wrote the correct formula, so why not use it?!!!!
The units do not work out either. (km/hr)*km = (km)^2/hr. What is that?! It certainly is not a unit of time!

Now (km)/(km/hr) = (km)*(hr/km) = hr, which IS a unit of time. Use units as they can be your best friend if you allow them to be.

 

[Marcus Clayson]

Dear Steven G

Thank you for your reply. I am working at the moment so don't have much time (!) or I would have replied sooner.
I do not know what Charles Darwin said about mathematicians, or, if he did write the statement you attributed to him, what he meant by it. It could be a reference to what those of us who wish we were mathematicians assume mathematics to be able to do, at least in part, given our existing levels of knowledge - find solutions to the conundrums of the material world, a journey without maps as it were, where each new development is built upon existing knowledge, the application of theories, new conclusions and hypotheses, which can then be tested and be found to be either valid or not. The assumption could be made, with reference to the above quotation, that absolute knowledge is unattainable, which is, of course, not an issue - hence the black cat is not there. I wonder if Shroedinger was of a similar mind...?

Anyway, getting back to the terrain of the mere mortal and my problem with linear equations, as a literacy specialist and teacher, my experience of mathematicians has led me to assume that the ability to communicate clearly is a fundamental problem for them. Perhaps mathematicians brains are indeed "wired differently", which is why individuals tend to be either capable with the humanities or with mathematics, and rarely at both. Of course, the abstract nature of mathematics, especially in relation to physics, and the inability of mathematicians to use a common lexis compound the problems. (How many mathematicians realise that there is a difference between square centimetres and centimetres squared?)

Thus your response to my question, where am I going wrong, failed entirely to answer the question - I am none the wiser. This is a state that I have found myself in so many times in the past, and which so many of my Learners have had to face during their encounters with mathematicians also. I repeat the mantra that I use every day in my teaching - what you have got to do is to answer the question.

As I see it, the problem is to find the distances (IN THE PLURAL!) travelled by 4 cars. I know the equation distance equals speed multiplied by time - i.e. d=s x t. What I did initially was assume that car A travels x kilometres at 40 kph, and car B travels y kilometres at 50kph. By a process of guesswork - i.e. NOT a mathematical method - I calculated that if car A travelled for 1 hour it would travel 40km and car B would travel 1.5 hours at 50kph and would thus travel 75 km. These are the correct solutions to the question, but they are arrived at by guesswork, NOT by the construction and solution of a pair of linear equations to be solved simultaneously, which was the form of the original task.

What I then assumed was that to set up the equations, I would multiply the distance travelled by car A - x km - by the time, then add the distance travelled by car B - y km - and do the same for the second set of values, thus creating two linear equations with two unknowns. I would then eliminate x or y, solve for x or y and then solve for the other missing variable. I have indeed constructed the equations assuming that speed x distance = time, which is erroneous, but that is the point of the thread - asking someone to EXPLAIN in clear, simple ENGLISH how I construct equations of this form WITHOUT THAT ERROR. The problem for me is that we do not know the times (in the singular) only the two times added together, which, in my equations, is another error. I am aware of this. So I repeat my original question: how do I set up the equations so that I can solve for the distances x and y without making these errors? I cannot see how to do it.

 

[Dr.Peterson]

What I then assumed was that to set up the equations, I would multiply the distance travelled by car A - x km - by the time, then add the distance travelled by car B - y km - and do the same for the second set of values, thus creating two linear equations with two unknowns. I would then eliminate x or y, solve for x or y and then solve for the other missing variable. I have indeed constructed the equations assuming that speed x distance = time, which is erroneous, but that is the point of the thread - asking someone to EXPLAIN in clear, simple ENGLISH how I construct equations of this form WITHOUT THAT ERROR. The problem for me is that we do not know the times (in the singular) only the two times added together, which, in my equations, is another error. I am aware of this. So I repeat my original question: how do I set up the equations so that I can solve for the distances x and y without making these errors? I cannot see how to do it.

The main issue here is your incorrect assumption that distance times time gives speed, as you also said here:

​

The correct equation, using your variables, is d = st, or s = d/t. (We often use r for rate, in part because "s" is too easily misread. Then we memorize d = rt, distance equals rate times time.)

This formula is very easily remembered in terms of units: The speed is measured in km/h, which is a distance (km) divided by a time (h), so s = d/t.

You can also think of d = st as "km = km/h * h", where the hours in denominator and numerator cancel.

When you correct this, you find that the time for d = x km at s = 40 km/h is t = d/s = x/40; and the time for d = y km at s = 50 km/h is t = d/s = y/50. So the first equation should be x/40 + y/50 = 2.5.

With the correct equations, you should be able to solve the problem algebraically.

 

[Marcus Clayson]

Dear Steven G and Subhotosh

Please accept my apologies for another late reply - I have been very busy at work. Thank you for taking the time to try to explain the problem: I am, however, none the wiser.
As I dislike trying to write mathematics using "Word", I have handwritten and photographed my response. The running order for the photographs is
DSCF0867, ...68...69
I know, I should update my software, but have neither the time, the money nor the inclination. I always think fondly of my hero, Isambard Kingdom Brunel and his legacy - and did he have a PC or a Mac? Did Euler, Newton, Einstein? Fancy modern fads do not make a difference - it's the ideas that count, pencil and paper...

 

[Dr.Peterson]

I think you copied a number wrong. Is it 50, as here,

​

or 40, as here?

​

If you make that 50, you get a nice answer.

But also, in your work, this doesn't look right:

​

You multiplied both numerator and denominator of the fractions by 4, which isn't how you multiply the fraction by 4; multiply only the numerator, since 4 is 4/1, not 4/4.

Similarly, in

​

are you saying the two coefficients of y are equal? They aren't.

Just be more careful, and you should get the answer.

 

[JeffM]

I find myself perplexed here. After getting a horrible crick in my neck trying to read images posted sideways, I still do not see a picture of the original problem.

It is frequently the case that students have difficulty with a problem because they misinterprest what the problem says. As a result, the students' paraphrases become misleading. I shall assume, however, that your most recent picture is an accurate paraphrase.

I currently also post at English Language Learners Stack Exchange, where I am currently rated in the top 1% in the world. So, I think I am competent to speak about literacy and communications. There is a technical vocabulary used by English speaking mathematicians, just as virtually all trades and disciplines have a technical vocabulary. It is true that many of the words in this technical vocabulary are words with a broader or different field of meaning in the general language. All this means is that when reading a mathematical text, you must construe words according to their meaning in mathematics. However, I greatly doubt that the technical vocabulary needed to understand the level of mathematics that you are currently studying is at all extensive or, with one or two exceptions, at all abstruse. I shall identify words in that technical vocabulary using italics when first used.

However, mathematics is done using a mathematical language called mathematical notation. To do mathematics, you must learn that language. When you talk about "setting up" equations, you are talking about translating from English into mathematical notation, nothing more. As someone who deals in language, you should recognize that translation involves knowing the grammar of the language into which you are translating. That grammar is the laws of algebra.

I like to start with a bilingual dictionary, as you started to do. And, when constructing my dictionary, I take care (i) to determine how many unknown numbers I am dealing with, and (ii) to assign a unique letter as a name to that number.

[math]x = \text {distance travelled by motorist A at 40 km/hr.\\ p = \text {hours travelled by motorist A at average speed of 40 km/hr.\\ y = \text {distance travelled by motorist A at average speed of 50 km/hr.\\ q = \text {hours travelled by motorist A at average speed of 50 km/hr.\\ u = \text {distance travelled by motorist B at average speed of 30 km/hr.\\ r = \text {hours travelled by motorist B at average speed of 30 km/hr.\\ v = \text {distance travelled by motorist B at average speed of 50 km/hr.\\ s = \text {hours travelled by motorist B at average speed of 50 km/hr.\\ [/math]
There are eight unknown numbers so I need eight equations. (To find n unknown numbers using elementary algebra, we always need n equations.) An equation states in mathematical notation that two mathematical expressions represent the same thing, the same quantity (number) in the case of elementary algebra. I also need to know a definition, namely that average speed = distance travelled [imath]\div[/imath] time travelled.

So I look for eight statements that say two QUANTITIES are the same. Here is what I get.

[math] \dfrac{x}{p} = 40;\\ \dfrac{y}{q} = 50;\\ \dfrac{u}{r} = 30;\\ \dfrac{v}{s} = 50;\\ p + q = 2.5;\\ r + s = 14;\\ u = 6x; \text { and}\\ v = 4y. [/math]
Do you see where each came from?

All I have done is to translate statements in the given problem (mostly in English) into mathematical notation and remind myself of a definition, which itself translates English words half-way into mathematical notation. There is nothing mysterious about it.

What I want to do next is to use the definition of average speed, the rules of algebra, and my eight equations to eliminate one by one the letters (pronumerals, meaning a symbol that stands for an unknown or unspecified number) until I reach an equation that equates a pronumeral to one or more numerals (a numeral being a symbol that stands for a unique number). This is a mechanical exercise that simply requires a systematic process. I am asked to find x and y so I want to eliminate everything except x and y.

[math]u = 6x \text { and } \dfrac{u}{r} = 30 implies \dfrac{6x}{r} = 30 \implies r = \dfrac{6x}{30} = \dfrac{x}{5}.\\ v = 4y \text { and } \dfrac{v}{s} = 50 implies \dfrac{4y}{s} = 50 \implies s = \dfrac{4y}{50} = \dfrac{2y}{25}.[/math]
I have eliminated u and v from consideration.

[math]r + s = 14, \ r = \dfrac{x}{5}, \text { and } s = \dfrac{2y}{25} \implies\\ 14 = \dfrac{x}{5} + \dfrac{2y}{25} = \dfrac{5x}{25} + \dfrac{2y}{25} = \dfrac{5x + 2y}{25} \implies\\ 5x + 2y = 25 * 14 = =350 .[/math]
I have eliminated r and s from consideration.

[math]\dfrac{x}{p} = 40 \implies p = \dfrac{x}{40.\\ \dfrac{y}{q} = 50 \implies q = \dfrac{y}{50}. \therefore p + q = 2.5 \implies \dfrac{x}{40} + \dfrac{y}{50} = 2.5 \implies\\ 2.5 = \dfrac{5x}{5 * 40} + \dfrac{4y}{4 * 50} = \dfrac{5x + 4y}{200} \implies\\ 5x + 4y = 2.5 * 200 = 500.[/math]
We know have two equations in x and y. Solve those and you have your answer.

We identified and counted the number of unknown quantities. We assigned a letter (a pronumeral) to each unknown quantity. We translated quantitative relationships expressed in English into equations in mathematical notation. We checked that we had the same number of equations as unknown quantities. And then we systematically employed the laws of algebra to reduce the number of pronumerals through our equations.
If you complete the last step of the process you should have your answer.

What is always hard is the translation from a natural language like English to the artificial language called mathematical notation. But that is what applied mathematics is all about. The rest is being systematic and knowing the grammar of your new language.

 

[Deleted member 4993]

- what you have got to do is to answer the question.

I disagree. The mantra should be:

what you have got to do is to HELP the student - to DISCOVER the answer through their own dint.​

Fancy modern fads do not make a difference - it's the ideas that count, pencil and paper...

Missed at least one other component - thinking through and application of common sense.

Common sense like:

speed = (distance) / (time) ............or...................distance = (speed) * (time) ............or.................. time = (distance) / (speed)

You are right - Fancy modern fads do (may) not make a difference

However. your ideas were garbled up here

Isambard Kingdom Brunel and his legacy - and did he have a PC or a Mac

And the builders of Pyramids or Taj Mahal did not have theodolite or dynamite and they did not complain about it.........

I have indeed constructed the equations assuming that speed x distance = time, which is erroneous, but that is the point of the thread - asking someone to EXPLAIN in clear, simple ENGLISH how I construct equations of this form

Very carefully and use common sense - no fads needed. The definition (equation) you should have known is:

\(\displaystyle (speed)\ =\ \frac{(distance)}{(time)}\)................... If you multiply by (time) on both sides you should get:

\(\displaystyle (speed) \ * \ \ (time) =\ \frac{(distance)}{(time)}\ * \ \ (time) \)......................then

\(\displaystyle (speed) \ * \ \ (time) =\ \frac{(distance)}{\cancel{(time)}}\ * \ \ \cancel{(time)} \).......................finally

\(\displaystyle (distance) \ = (speed) \ * \ (time)\)

Nothing more than middle school algebra and common sense - no fads.

where am I going wrong, failed entirely to answer the question

Was answered - you failed to READ it

The main reason your equations are wrong is because t ≠ r*d !!!

as a literacy specialist and teacher

Really .......teach yourself a bit of "reading comprehension"

 

[JeffM]

I find myself perplexed here. After getting a horrible crick in my neck trying to read images posted sideways, I still do not see a picture of the original problem.

It is frequently the case that students have difficulty with a problem because they misinterprest what the problem says. As a result, the students' paraphrases become misleading. I shall assume, however, that your most recent picture is an accurate paraphrase.

I currently also post at English Language Learners Stack Exchange, where I am currently rated in the top 1% in the world. So, I think I am competent to speak about literacy and communications. There is a technical vocabulary used by English speaking mathematicians, just as virtually all trades and disciplines have a technical vocabulary. It is true that many of the words in this technical vocabulary are words with a broader or different field of meaning in the general language. All this means is that when reading a mathematical text, you must construe words according to their meaning in mathematics. However, I greatly doubt that the technical vocabulary needed to understand the level of mathematics that you are currently studying is at all extensive or, with one or two exceptions, at all abstruse. I shall identify words in that technical vocabulary using italics when first used.

However, mathematics is done using a mathematical language called mathematical notation. To do mathematics, you must learn that language. When you talk about "setting up" equations, you are talking about translating from English into mathematical notation, nothing more. As someone who deals in language, you should recognize that translation involves knowing the grammar of the language into which you are translating. That grammar is the laws of algebra.

I like to start with a bilingual dictionary, as you started to do. And, when constructing my dictionary, I take care (i) to determine how many unknown numbers I am dealing with, and (ii) to assign a unique letter as a name to that number.

[math]x = \text {distance travelled by motorist A at 40 km/hr.\\ p = \text {hours travelled by motorist A at average speed of 40 km/hr.\\ y = \text {distance travelled by motorist A at average speed of 50 km/hr.\\ q = \text {hours travelled by motorist A at average speed of 50 km/hr.\\ u = \text {distance travelled by motorist B at average speed of 30 km/hr.\\ r = \text {hours travelled by motorist B at average speed of 30 km/hr.\\ v = \text {distance travelled by motorist B at average speed of 50 km/hr.\\ s = \text {hours travelled by motorist B at average speed of 50 km/hr.\\ [/math]
There are eight unknown numbers so I need eight equations. (To find n unknown numbers using elementary algebra, we always need n equations.) An equation states in mathematical notation that two mathematical expressions represent the same thing, the same quantity (number) in the case of elementary algebra. I also need to know a definition, namely that average speed = distance travelled [imath]\div[/imath] time travelled.

So I look for eight statements that say two QUANTITIES are the same. Here is what I get.

[math] \dfrac{x}{p} = 40;\\ \dfrac{y}{q} = 50;\\ \dfrac{u}{r} = 30;\\ \dfrac{v}{s} = 50;\\ p + q = 2.5;\\ r + s = 14;\\ u = 6x; text { and}\\ v = 4y. [/math]
Do you see where each came from?

All I have done is to translate statements in the given problem (mostly in English) into mathematical notation and remind myself of a definition, which itself translates English words half-way into mathematical notation. There is nothing mysterious about it.

What I want to do next is to use the definition of average speed, the rules of algebra, and my eight equations to eliminate one by one the letters (pronumerals, meaning a symbol that stands for an unknown or unspecified number) until I reach an equation that equates a pronumeral to one or more numerals (a numeral being a symbol that stands for a unique number). This is a mechanical exercise that simply requires a systematic process. I am asked to find x and y so I want to eliminate everything except x and y.

[math]u = 6x \text { and } \dfrac{u}{r} = 30 implies \dfrac{6x}{r} = 30 \implies r = \dfrac{6x}{30} = \dfrac{x}{5}.\\ v = 4y \text { and } \dfrac{v}{s} = 50 implies \dfrac{4y}{s} = 50 \implies s = \dfrac{4y}{50} = \dfrac{2y}{25}.[/math]
I have eliminated u and v from consideration.

[math]r + s = 14, \ r = \dfrac{x}{5}, \text { and } s = \dfrac{2y}{25} \implies\\ 14 = \dfrac{x}{5} + \dfrac{2y}{25} = \dfrac{5x}{25} + \dfrac{2y}{25} = \dfrac{5x + 2y}{25} \implies\\ 5x + 2y = 25 * 14 = =350 .[/math]
I have eliminated r and s from consideration.

[math]\dfrac{x}{p} = 40 \implies p = \dfrac{x}{40.\\ \dfrac{y}{q} = 50 \implies q = \dfrac{y}{50}. \therefore p + q = 2.5 \implies \dfrac{x}{40} + \dfrac{y}{50} = 2.5 \implies\\ 2.5 = \dfrac{5x}{5 * 40} + \dfrac{4y}{4 * 50} = \dfrac{5x + 4y}{200} \implies\\ 5x + 4y = 2.5 * 200 = 500.[/math]
We know have two equations in x and y. Solve those and you have your answer.

We identified and counted the number of unknown quantities. We assigned a letter (a pronumeral) to each unknown quantity. We translated quantitative relationships expressed in English into equations in mathematical notation. We checked that we had the same number of equations as unknown quantities. And then we systematically employed the laws of algebra to reduce the number of pronumerals through our equations.
If you complete the last step of the process you should have your answer.

What is always hard is the translation from a natural language like English to the artificial language called mathematical notation. But that is what applied mathematics is all about. The rest is being systematic and knowing the grammar of your new language.

I see that there were two typographical errors in my earlier post. Here is what should have been rendered.

[math]\dfrac{x}{p} = 40 \implies p = \dfrac{x}{40}.\\ \dfrac{y}{q} = 50 \implies q = \dfrac{y}{50}.\\ \therefore p + q = 2.5 \implies \dfrac{x}{40} + \dfrac{y}{50} = 2.5 \implies\\ 2.5 = \dfrac{5x}{5 * 40} + \dfrac{4y}{4 * 50} = \dfrac{5x + 4y}{200} \implies\\ 5x + 4y = 2.5 * 200 = 500.[/math]
We now have two equations in x and y. Solve those and you have your answer.

 

[JeffM]

I find myself perplexed here. After getting a horrible crick in my neck trying to read images posted sideways, I still do not see a picture of the original problem.

It is frequently the case that students have difficulty with a problem because they misinterprest what the problem says. As a result, the students' paraphrases become misleading. I shall assume, however, that your most recent picture is an accurate paraphrase.

I currently also post at English Language Learners Stack Exchange, where I am currently rated in the top 1% in the world. So, I think I am competent to speak about literacy and communications. There is a technical vocabulary used by English speaking mathematicians, just as virtually all trades and disciplines have a technical vocabulary. It is true that many of the words in this technical vocabulary are words with a broader or different field of meaning in the general language. All this means is that when reading a mathematical text, you must construe words according to their meaning in mathematics. However, I greatly doubt that the technical vocabulary needed to understand the level of mathematics that you are currently studying is at all extensive or, with one or two exceptions, at all abstruse. I shall identify words in that technical vocabulary using italics when first used.

However, mathematics is done using a mathematical language called mathematical notation. To do mathematics, you must learn that language. When you talk about "setting up" equations, you are talking about translating from English into mathematical notation, nothing more. As someone who deals in language, you should recognize that translation involves knowing the grammar of the language into which you are translating. That grammar is the laws of algebra.

I like to start with a bilingual dictionary, as you started to do. And, when constructing my dictionary, I take care (i) to determine how many unknown numbers I am dealing with, and (ii) to assign a unique letter as a name to that number.

[math]x = \text {distance travelled by motorist A at 40 km/hr.\\ p = \text {hours travelled by motorist A at average speed of 40 km/hr.\\ y = \text {distance travelled by motorist A at average speed of 50 km/hr.\\ q = \text {hours travelled by motorist A at average speed of 50 km/hr.\\ u = \text {distance travelled by motorist B at average speed of 30 km/hr.\\ r = \text {hours travelled by motorist B at average speed of 30 km/hr.\\ v = \text {distance travelled by motorist B at average speed of 50 km/hr.\\ s = \text {hours travelled by motorist B at average speed of 50 km/hr.\\ [/math]
There are eight unknown numbers so I need eight equations. (To find n unknown numbers using elementary algebra, we always need n equations.) An equation states in mathematical notation that two mathematical expressions represent the same thing, the same quantity (number) in the case of elementary algebra. I also need to know a definition, namely that average speed = distance travelled [imath]\div[/imath] time travelled.

So I look for eight statements that say two QUANTITIES are the same. Here is what I get.

[math] \dfrac{x}{p} = 40;\\ \dfrac{y}{q} = 50;}\\ \dfrac{u}{r} = 30;}\\ \dfrac{v}{s} = 50;}\\ p + q = 2.5;}\\ r + s = 14;}\\ u = 6x; text { and}\\ v = 4y. [/math]
Do you see where each came from?

All I have done is to translate statements in the given problem (mostly in English) into mathematical notation and remind myself of a definition, which itself translates English words half-way into mathematical notation. There is nothing mysterious about it.

What I want to do next is to use the definition of average speed, the rules of algebra, and my eight equations to eliminate one by one the letters (pronumerals, meaning a symbol that stands for an unknown or unspecified number) until I reach an equation that equates a pronumeral to one or more numerals (a numeral being a symbol that stands for a unique number). This is a mechanical exercise that simply requires a systematic process. I am asked to find x and y so I want to eliminate everything except x and y.

[math]u = 6x \text { and } \dfrac{u}{r} = 30 implies \dfrac{6x}{r} = 30 \implies r = \dfrac{6x}{30} = \dfrac{x}{5}.\\ v = 4y \text { and } \dfrac{v}{s} = 50 implies \dfrac{4y}{s} = 50 \implies s = \dfrac{4y}{50} = \dfrac{2y}{25}.[/math]
I have eliminated u and v from consideration.

[math]r + s = 14, \ r = \dfrac{x}{5}, \text { and } s = \dfrac{2y}{25} \implies\\ 14 = \dfrac{x}{5} + \dfrac{2y}{25} = \dfrac{5x}{25} + \dfrac{2y}{25} = \dfrac{5x + 2y}{25} \implies\\ 5x + 2y = 25 * 14 = =350 .[/math]
I have eliminated r and s from consideration.

[math]\dfrac{x}{p} = 40 \implies p = \dfrac{x}{40.\\ \dfrac{y}{q} = 50 \implies q = \dfrac{y}{50}. \therefore p + q = 2.5 \implies \dfrac{x}{40} + \dfrac{y}{50} = 2.5 \implies\\ 2.5 = \dfrac{5x}{5 * 40} + \dfrac{4y}{4 * 50} = \dfrac{5x + 4y}{200} \implies\\ 5x + 4y = 2.5 * 200 = 500.[/math]
We know have two equations in x and y. Solve those and you have your answer.

We identified and counted the number of unknown quantities. We assigned a letter (a pronumeral) to each unknown quantity. We translated quantitative relationships expressed in English into equations in mathematical notation. We checked that we had the same number of equations as unknown quantities. And then we systematically employed the laws of algebra to reduce the number of pronumerals through our equations.
If you complete the last step of the process you should have your answer.

What is always hard is the translation from a natural language like English to the artificial language called mathematical notation. But that is what applied mathematics is all about. The rest is being systematic and knowing the grammar of your new language.

There was another coding error in my original post.

[math]x = \text {distance travelled by motorist A at 40 km/hr.}\\ p = \text {hours travelled by motorist A at average speed of 40 km/hr}.\\ y = \text {distance travelled by motorist A at average speed of 50 km/hr.}\\ q = \text {hours travelled by motorist A at average speed of 50 km/hr.}\\ u = \text {distance travelled by motorist B at average speed of 30 km/hr.}\\ r = \text {hours travelled by motorist B at average speed of 30 km/hr.}\\ v = \text {distance travelled by motorist B at average speed of 50 km/hr.}\\ s = \text {hours travelled by motorist B at average speed of 50 km/hr.}\\ [/math]

 

[Marcus Clayson]

First, 1000 apologies for the length of time it has taken me to respond to all of the comments, which were both helpful and unhelpful, but all gratefully received.

I did finally work out how to solve the problem, and I am attaching a photograph of my working if anyone is interested. I had an inkling about how to solve it, but have never, in all of my days, seen a set of linear equations constructed from other formulae. I was reminded of the apocryphal saying, that if a chimpanzee were sat down in front of a typewriter, it would eventually turn out the works of Shakespeare. Prejudices about Shakespeare's work aside, this is, of course, a complete nonsense, but I feel that if I had been sat down in front of a computer and left to solve the problem (or relied on some of the responses I received), I would never have got it. I hope, however, that this knowledge/experience is something that I can carry forward.

So, again, many thanks for the support, and for the help some of you gave. I shall, no doubt come across another problem in the future, the solution to which is simple and logical, but which eludes me, and shall again, humbly ask for guidance, and shall, next time, hopefully, show more restraint, panic less...

Best wishes to you all.