Forming two equations

[confusionwithconfucius]

I've had some trouble with this question and turning it into two equations. Question is:

"There were 96 people in a party. One man leaves and is replaced by a woman, leaving three times as many women as men at the party.
a) Form two equations of the above information given that m represents men and w represents women at the party."

I'm not good at forming equations from worded questions and am having trouble seeing a solution here and it's driving me up the wall. The second part of the question asks that each equation be solved simultaneously and I feel adequately confident to do that. It's just forming the equations that has me stumped, so any assistance would be greatly appreciated.

I know it's not an equation, but could the figures be resolved by dividing 96 by 3, giving you 32 as the total number of men and then subtracting 32 from 96 to give the total number of women at 64?

Thanks in advance.

 

[confusionwithconfucius]

I know it's not an equation, but could the figures be resolved by dividing 96 by 3, giving you 32 as the total number of men and then subtracting 32 from 96 to give the total number of women at 64?

Rather, in particular to this bit using this method, would the two equations be: m=96/3 and w=96-m? Which you could then solve and substitute into w+m=96?

I'll just include what the 2nd part of the question is which I should have included originally: "b) Solve the two equation simultaneously to find how many men and women were there originally at the party?"

 

[lex]

Draw a 'picture' and write an equation for each:

 

[Harry_the_cat]

Ok let's test out your answer. You say there are 32 men and 64 women.

Are there 96 people at the party? Does 32+64 = 96 Yes

If one man leaves (now 31 men) and is replaced by a woman (now 65 women), are there three times as many women as men at the party.
Does 65 = 3 x 31? No

SO YOUR SOLUION IS INCORRECT.

Here's the steps you need to go through:

1. If there are 96 people in total and there are m men and w women, what does m + w = ? That's the first equation.

2. If there are m men and one man leaves, how many men are there now (in terms of m)?

3. If there are w women and one more woman comes in, how many women are there now (in terms of w)?

4. There are now "three times as many women as men". Can you use your answer to 2 and 3 to form an equation for this information?

You will then have 2 equations (one from 1 and the other from 4).

Show us your equations first.

 

[confusionwithconfucius]

Ok let's test out your answer. You say there are 32 men and 64 women.

Are there 96 people at the party? Does 32+64 = 96 Yes

If one man leaves (now 31 men) and is replaced by a woman (now 65 women), are there three times as many women as men at the party.
Does 65 = 3 x 31? No

SO YOUR SOLUION IS INCORRECT.

Here's the steps you need to go through:

1. If there are 96 people in total and there are m men and w women, what does m + w = ? That's the first equation.

2. If there are m men and one man leaves, how many men are there now (in terms of m)?

3. If there are w women and one more woman comes in, how many women are there now (in terms of w)?

4. There are now "three times as many women as men". Can you use your answer to 2 and 3 to form an equation for this information?

You will then have 2 equations (one from 1 and the other from 4).

Show us your equations first.

I'm still not understanding.

m + w has to equal 96 right? So the first equation has to be m + w = 96

Now we can say that, yes, 32+64=96 (with the number of woman being triple in size compared to the men) but this is not the original number of people at the party. These are the figures after one man has left and one woman has arrived. So with that, could we not assume the original amount of people at the party is 33 men and 63 women which still totals 96, however the amount of women is not 3 times the men at this point as 33x3=99 which can't be true as there is only 96 people. But after 1 man has left and 1 woman has arrived, it still totals to 96 people, however now it is true that there are 3 times as many women as there are men?

So there is obviously a m - 1 and w + 1 here but I just don't understand how to make this into an equation to be solved. I assume that the second part of the question is basically saying that this will be a simultaneous linear equation to solve, or am I perhaps getting to hung up on the terminology, assuming it to be something that it isn't?

 

[Harry_the_cat]

Yes m+w = 96. That's the first equation. Leave that for now.

Ok after 1 man leaves there are m-1 men.
When 1 woman joins, there are w+1 women.
At the stage there are 3 times as many women as men,
so w+1 = ? x (m-1).

 

[JeffM]

I'm still not understanding.

m + w has to equal 96 right? So the first equation has to be m + w = 96

Now we can say that, yes, 32+64=96 (with the number of woman being triple in size compared to the men) but this is not the original number of people at the party. These are the figures after one man has left and one woman has arrived. So with that, could we not assume the original amount of people at the party is 33 men and 63 women which still totals 96, however the amount of women is not 3 times the men at this point as 33x3=99 which can't be true as there is only 96 people. But after 1 man has left and 1 woman has arrived, it still totals to 96 people, however now it is true that there are 3 times as many women as there are men?

So there is obviously a m - 1 and w + 1 here but I just don't understand how to make this into an equation to be solved. I assume that the second part of the question is basically saying that this will be a simultaneous linear equation to solve, or am I perhaps getting to hung up on the terminology, assuming it to be something that it isn't?

I also do not understand, but in my case I do not understand your confusion.

I like to start by identifying how many numbers are initially unknown. And I must admit that the problem as stated is ambiguous.
Let’s fix that.

$$m = \text {number of men at the party before anyone leaves.}\\ n = \text {number of men at the party after first person leaves.}\\ w = \text {number of women at the party before anyone leaves.}\\ x = \text {number of women at the party after person arrives very late.}$$
Can you see that the problem is ambiguous in that we cannot be sure whether we are being asked about the number of men and women before or after the change? Consequently, we are looking for four numbers. We have assigned a distinct letter to each unknown. We have four unknowns. That means a solution requires four equations. (To guarantee a solution further requires that those equations be independent and consistent.)

Now we must translate from English to math.

“There were 96 people at a party” $\implies m + w = 96.$

“One man leaves” $\implies n = m - 1.$

”and is replaced by a woman“ $\implies x = w + 1.$

”leaving three times as many woman as men” \implies $\implies 3n = x.$

The real difficulty in translation is in fully understanding the English. And do you see how much easier it is if every unknown number is given its own letter? For some IDIOTIC reason, the normal way to teach is to minimize the number of letters and make students solve admittedly easy algebra problems in their heads before writing anything down.

Now we have four equations in four unknowns. The general way to solve these systems is to express one unknown in terms of one or more of the other unknowns and then replace that unknown in all the other equations. We keep doing that until we have one equation in one unknown and solve that through basic algebra. In this case, it is really easy because two of our equations already express one unknown in terms of another.

$$n = m - 1 \text { and } 3n = x \implies 3(m - 1) = x \implies 3m - 3 = x.$$
$$x = w + 1 \text { and } 3m - 3 = x \implies 3m - 3 = w + 1 \implies w = 3m - 4.$$
We have now reduced our system to two equations in two unknowns, namely

$$w + m = 96 \text { and } 3m - 4 = w.$$
Now complete the process of reducing to one equation in one unknown and solve that last equation.That gives you one of the unknown numbers. Now go back and solve for the remaining numbers.

What harry told you was how to check your answers, not find them.

 

[Harry_the_cat]

@JeffM I agree that the question could be seen as somewhat ambiguous, but I think it is meant to be a lot simpler than what you have suggested. The original poster is obviously a beginner learning about simultaneous equations. I don't think it is idiotic to think only two variables need to be involved.

 

[JeffM]

@JeffM I agree that the question could be seen as somewhat ambiguous, but I think it is meant to be a lot simpler than what you have suggested. The original poster is obviously a beginner learning about simultaneous equations. I don't think it is idiotic to think only two variables need to be involved.

I recognize that simultaneous equations are usually introduced late in the game. I believe that is entirely wrong pedagogically. Virtually no real world problem requires finding a single number. Practical applications of algebra usually require finding many numbers.

Moreover, the traditional way of teaching algebra requires setting up expressions for some unknowns without explicitly acknowledging that process. I have found it much easier for students to count and discretely label every unknown but relevant number. In most cases, it is then easy to translate a natural language into mathematical notation.

Of course, this is just opinion, but I think labeling unknowns through complex expressions seems simple only to those who already “get algebra.”

 

[Steven G]

I also do not understand, but in my case I do not understand your confusion.

I like to start by identifying how many numbers are initially unknown. And I must admit that the problem as stated is ambiguous.
Let’s fix that.

$$m = \text {number of men at the party before anyone leaves.}\\ n = \text {number of men at the party after first person leaves.}\\ w = \text {number of women at the party before anyone leaves.}\\ x = \text {number of women at the party after person arrives very late.}$$
Can you see that the problem is ambiguous in that we cannot be sure whether we are being asked about the number of men and women before or after the change? Consequently, we are looking for four numbers. We have assigned a distinct letter to each unknown. We have four unknowns. That means a solution requires four equations. (To guarantee a solution further requires that those equations be independent and consistent.)

Now we must translate from English to math.

“There were 96 people at a party” $\implies m + w = 96.$

“One man leaves” $\implies n = m - 1.$

”and is replaced by a woman“ $\implies x = w + 1.$

”leaving three times as many woman as men” \implies $\implies 3n = x.$

The real difficulty in translation is in fully understanding the English. And do you see how much easier it is if every unknown number is given its own letter? For some IDIOTIC reason, the normal way to teach is to minimize the number of letters and make students solve admittedly easy algebra problems in their heads before writing anything down.

Now we have four equations in four unknowns. The general way to solve these systems is to express one unknown in terms of one or more of the other unknowns and then replace that unknown in all the other equations. We keep doing that until we have one equation in one unknown and solve that through basic algebra. In this case, it is really easy because two of our equations already express one unknown in terms of another.

$$n = m - 1 \text { and } 3n = x \implies 3(m - 1) = x \implies 3m - 3 = x.$$
$$x = w + 1 \text { and } 3m - 3 = x \implies 3m - 3 = w + 1 \implies w = 3m - 4.$$
We have now reduced our system to two equations in two unknowns, namely

$$w + m = 96 \text { and } 3m - 4 = w.$$
Now complete the process of reducing to one equation in one unknown and solve that last equation.That gives you one of the unknown numbers. Now go back and solve for the remaining numbers.

What harry told you was how to check your answers, not find them.

Jeff, This time you are wrong, in my opinion. I think that the question is saying that as soon as one man leaves, a woman replaces this man.

 

[JeffM]

Jeff, This time you are wrong, in my opinion. I think that the question is saying that as soon as one man leaves, a woman replaces this man.

Steven I agree about the replacement.

What is ambiguous is whether the problem is asking for the number of men and the number of women before or after the replacement.

 

[The Highlander]

It's a brave man indeed that is (fool)hardy enough to join in this argument. ?

But I have to say that I agree wholeheartedly with Jeff that the question is thoroughly ambiguous in that it is not clear whether the m & w are intended to represent the number of men & women (respectively) who are present before or after the man who leaves is replaced by a woman.

However, I also agree (just as wholeheartedly) with Harry & Steven that the question setter fully intended that the m & w were supposed to represent the original number of each sex present, hence, my vote goes to Harry's method! ??

 

[JeffM]

I want to stress four things.

First, the issue of whether this specific problem is ambiguous in not defining what is to be found is distinct from what is the best generic method to introduce algebraic word problems to students.

Second, I wholeheartedly agree with harry that checking the answer against the original equations is essential. I have no disagreement there.

Third, I also agree that harry’s method of solving this specific problem is perfectly valid, is natural for anyone already adept at algebra, and is what is typically taught. I am not objecting to the method itself. I am objecting to teaching it as an introduction to word problems.

Fourth, most students find learning algebra to be hard. Virtually everyone I have tutored in algebra, even those quite competent in the mechanics of algebra, struggle with “word problems.” But if you cannot do word problems, you cannot apply algebra to anything. We should be looking for methods to teach algebra, particularly word problems, that are as obvious to beginning students as possible.

There are four unknown numbers. Harry lists four facts. Then Harry says that immediately gives us two equations in two unknowns. The student says “What? So you mean four quantitative facts and four unknowns immediately leads to two equations in two unknowns?” This of course is the wrong inference to make.

What is generally correct is that with n unknown numbers, we need at least n quantitative facts to find numerical values for the unknown numbers. Each of Harry’s listed facts leads to a simple equation. Two equations are so simple that those familiar with algebra can and will take a short cut and go immediately to two equations. But the short cut obscures the general process and will not work for more complicated problems. If a student is confused, using a short cut that apparently contradicts the general rule of “n unknowns require n equations” makes no sense.

Consider this analogy.

$$7x - 34 = 4x + 23 \implies\\ 7x - 34 + 34 = 4x + 23 + 34 \implies\\ 7x = 4x + 57 \implies \\ 7x - 4x = 4x - 4x + 57 \implies\\ 3x = 57 \implies \\ \dfrac{3x}{3} = \dfrac{57}{3} \implies \\ x = 19.$$
That is how we teach beginning students how to solve an equation. Later of course we teach short cuts like

$$ax + b = cx + d \implies (a - c)x = d - b.$$
But we do not start there.

The traditional method of teaching students how to solve word problems makes no sense. It is not general. It requires doing algebra in your head. But it seems obvious to anyone who already knows algebra.

Here is a general method.

(1) Identify the potentially relevant quantities that are unknown.
(2) Assign a distinct letter to each of those unknown quantities.
(3) Identify, either through information in the problem itself or from information everyone is expected to know, at least as many quantitative facts about the relations among the unknown quantities as there are unknown quantities.
(4) Turn those facts into equations if possible.
(5) Apply algebra to those equations.
(6) Check your answers against the original equations.

For word problems that are soluble by algebra, this method works every time.

But, in general, we do not teach this method consistently to beginning students before later showing them that some of the steps can sometimes be skipped. I view that general lack of consistency to be wrong pedagogically.

Now I must apologize. In context, it appears that I was calling Harry_the_Cat idiotic. That of course was not at all my intention. I admire her very much. In fact, I gave her first post in this thread a very prompt “like.”

Because I never was a professional teacher and only tutored before and then again after my years of employment, I never was taught how to teach algebra. I learned from one-on-one experience, mostly when I was only a few years older than my tutees. When teaching beginning algebra, I found that it helps a lot to forget much of what I knew about math and to remember how strange algebra seemed when I was twelve. What is obvious and thus simple to those of us comfortable with algebra and calculus is often far from obvious to a beginning student.

I think consistency is very important. Identify, label, and count every unknown. With n unknowns, look for n equations involving those unknowns.

 

[Steven G]

Jeff,
I read the problem again, English is not my thing, and now I see your point perfectly. Sorry for saying that I disagreed with you.

 

[JeffM]

Jeff,
I read the problem again, English is not my thing, and now I see your point perfectly. Sorry for saying that I disagreed with you.

Not a problem my friend. People can disagree and still respect each other: my sister and I have very different politics but get along just fine.

Moreover, disagreement can eventually be productive in a Hegelian way: thesis, antithesis, synthesis.

And finally some disagreements turn out not even to exist.

 

[Deleted member 4993]

Hegelian way

Hegel !!! That's where communism starts .....

 

[lex]

Hegel !!! That's where communism starts .....

That's exactly what came to my mind!! - but I didn't like to stoke anything further up.

 

[JeffM]

Well fortunately I said not a word about Marx. I was thinking more along the lines of Kant-Fichte-Hegel. Idea-critique-improved idea as a positive view of disagreement.