Linear Equations: teacher distributes $2 among 20 children...

Tadams052012

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Hi there,

This is my second post on this forum, I'm self studying Algebra from a book called: Teach Yourself Algebra: A Complete Introduction. Since the last problem you guys helped me with I've been cruising along and have reached 'linear equations' something I've been quite enjoying, until this question that is...

11. A teacher distributes $2 among 20 children, giving 5 cents to some and 25 cents each to the rest. How many children received 25 cents each?

Now my difficulty here is not solving the problem per se, I've done so using mental reasoning (a kind of trial and error method where I go through different combinations of fives and twenty-fives until they make up twenty with their total equaling two-hundred cents of course), the answer is five.

No, my problem is putting this conundrum into the form of a linear equation and solving it algebraically on paper. This is what I need to be able to do, since obviously, once the problems become more complex my patchy mental trial and error is no longer going to be reliable or even workable.

Is there anyone out there who might be able to take me through the process of solving this problem using a linear equation?

Just to illustrate my thought process and perhaps give you an inkling of where I'm going wrong: the first thing I'm doing is looking for the unknown to be ascertained, which is obviously the number of children receiving 25 cents, I assign this a letter - X.

Now where my problem begins is that there is also another unknown in this situation: the number of children who received 5 cents, and the two unknowns are interdependent, so it seems I must give this one a letter also - Y.

Now having two unknowns in my equation seems to throw up all kinds of problems since to find the value of one it needs to be isolated on one side with a numerical value on the other. This means removing the other of the two unknowns from the equation, which seems impossible at least as long as it remains in letter form i.e. subtract Y from one side means to subtract it from the other which means you then have -Y on the other side.

Here are the only things resembling linear equations I've been able to formulate:

X+Y=20

and

25X + 5Y = 200

Any help would be much appreciated.

Many thanks,

Tom.
 
Hi there,

This is my second post on this forum, I'm self studying Algebra from a book called: Teach Yourself Algebra: A Complete Introduction. Since the last problem you guys helped me with I've been cruising along and have reached 'linear equations' something I've been quite enjoying, until this question that is...

11. A teacher distributes $2 among 20 children, giving 5 cents to some and 25 cents each to the rest. How many children received 25 cents each?

Now my difficulty here is not solving the problem per se, I've done so using mental reasoning (a kind of trial and error method where I go through different combinations of fives and twenty-fives until they make up twenty with their total equaling two-hundred cents of course), the answer is five.

No, my problem is putting this conundrum into the form of a linear equation and solving it algebraically on paper. This is what I need to be able to do, since obviously, once the problems become more complex my patchy mental trial and error is no longer going to be reliable or even workable.

Is there anyone out there who might be able to take me through the process of solving this problem using a linear equation?

Just to illustrate my thought process and perhaps give you an inkling of where I'm going wrong: the first thing I'm doing is looking for the unknown to be ascertained, which is obviously the number of children receiving 25 cents, I assign this a letter - X.

Now where my problem begins is that there is also another unknown in this situation: the number of children who received 5 cents, and the two unknowns are interdependent, so it seems I must give this one a letter also - Y.

Now having two unknowns in my equation seems to throw up all kinds of problems since to find the value of one it needs to be isolated on one side with a numerical value on the other. This means removing the other of the two unknowns from the equation, which seems impossible at least as long as it remains in letter form i.e. subtract Y from one side means to subtract it from the other which means you then have -Y on the other side.

Here are the only things resembling linear equations I've been able to formulate:

X+Y=20

and

25X + 5Y = 200

Any help would be much appreciated.

Many thanks,

Tom.
Your thought process and the two equations are correct.

You have two equations and two unknowns. But this is just the beginning. Later in your professional life you may encounter 1000 variables and 1000 equations.

Now go forth and multiply - solve those two unknowns.....
 
Hi there,

This is my second post on this forum, I'm self studying Algebra from a book called: Teach Yourself Algebra: A Complete Introduction. Since the last problem you guys helped me with I've been cruising along and have reached 'linear equations' something I've been quite enjoying, until this question that is...

11. A teacher distributes $2 among 20 children, giving 5 cents to some and 25 cents each to the rest. How many children received 25 cents each?

Now my difficulty here is not solving the problem per se, I've done so using mental reasoning (a kind of trial and error method where I go through different combinations of fives and twenty-fives until they make up twenty with their total equaling two-hundred cents of course), the answer is five.

No, my problem is putting this conundrum into the form of a linear equation and solving it algebraically on paper. This is what I need to be able to do, since obviously, once the problems become more complex my patchy mental trial and error is no longer going to be reliable or even workable.

Is there anyone out there who might be able to take me through the process of solving this problem using a linear equation?

Just to illustrate my thought process and perhaps give you an inkling of where I'm going wrong: the first thing I'm doing is looking for the unknown to be ascertained, which is obviously the number of children receiving 25 cents, I assign this a letter - X.

Now where my problem begins is that there is also another unknown in this situation: the number of children who received 5 cents, and the two unknowns are interdependent, so it seems I must give this one a letter also - Y.

Now having two unknowns in my equation seems to throw up all kinds of problems since to find the value of one it needs to be isolated on one side with a numerical value on the other. This means removing the other of the two unknowns from the equation, which seems impossible at least as long as it remains in letter form i.e. subtract Y from one side means to subtract it from the other which means you then have -Y on the other side.

Here are the only things resembling linear equations I've been able to formulate:

X+Y=20

and

25X + 5Y = 200

Any help would be much appreciated.

Many thanks,

Tom.
Your two equations are correct. I am not sure where your problem is. Have you solved two equations with two unknowns before? Are the substitution method, elimination method and cramer's rule known to you? If not, please follow these links.
https://www.youtube.com/watch?v=Qb5bV9c4z50&list=PLtmopxjCQWixtwGXPnM3dmyrDw0vHlThM&index=9
https://www.youtube.com/watch?v=AQRvwfg4fVs&list=PLtmopxjCQWixtwGXPnM3dmyrDw0vHlThM&index=10
https://www.youtube.com/watch?v=xJTqgZmcKQI&list=PLtmopxjCQWixtwGXPnM3dmyrDw0vHlThM&index=13

I would first divide the 2nd equation by 5, since 5 is a common factor on both sides of the equation
 
11. A teacher distributes $2 among 20 children, giving 5 cents to some and 25 cents each to the rest. How many children received 25 cents each?
Here are the only things resembling linear equations I've been able to formulate:
X+Y=20 and 25X + 5Y = 200
@Tadams, for self-studying you did a good job translating the statement into symbolic form.
As you do more of these I suggest that if you have an equation like your first the use substitution.
\(\displaystyle X+Y=20\) so \(\displaystyle Y=20-X\).
Using substitution \(\displaystyle 25X+5(20-X)=200\). Easy to solve?

Please tell us why you think that I chose \(\displaystyle Y\) to use in the substitution?
 
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I approach teaching how to solve word problems in a way that is completely different from standard pedagogy.

Algebra is about working with numbers that are not yet known. There are basically two aspects to algebra: (1) mechanical techniques for transforming expressions and solving equations, (2) finding the expressions and equations that apply to a particular problem. The second is what makes algebra useful; the first is what makes algebra possible.

It is standard to start with word problems that can be solved by using a single linear equation with one unknown. Because any solvable problem with only one number that is unknown can be solved by arithmetic, these initial algebra problems almost always involve more than one unknown number. To create a single equation involving two or more unknown numbers in terms of just one of the unknown numbers can take considerable mental agility on the part of a beginning student.

So standard pedagogy starts with a slight scam. You are taught how to solve linear equations with one unknown because such equations are the simplest to solve, but the word problems themselves involve more than one unknown number. You must somehow figure out how to express all the unknown numbers in terms of one of them. And then, as happened to you, you run into a problem where that trick is either impossible or you personally can't make it work.

When I tutor kids in algebra, I start on word problems by telling them to assign a variable (technically a pronumeral) to each number in the problem that is not yet known. This is far easier to do than figuring out how to express them all in terms of one pronumeral. I then tell them that to solve the problem algebraically they need to find as many valid equations as they have pronumerals. This is usually not too hard: each relevant detail given in words usually translates nicely into an equation.

So my way of teaching how to solve word problems immediately forces students to solve systems of linear equations. That means that they face from the beginning a somewhat more challenging mechanical problem, but one that can be immediately applied to a much wider range of word problems and that is easier when translating from words into math.

There are several ways to solve systems of linear equations. I stress the method of substitution, which is the most general. So, for your problem specifically, we have

\(\displaystyle x + y = 20 \text { and } 25x + 5y = 200.\)

We pick the simpliest equation and solve for one of the unknowns: this will be an algebraic solution rather than a numeric solution.

\(\displaystyle x + y = 20 \implies x = 20 - y.\)

Not difficult in this case. Now we substitute our new value for x into the remaining equation

\(\displaystyle 25x + 5y = 200 \implies 25(20 - y) + 5y = 200.\)

Now you have one linear equation in one unknown, namely y. Solve it numerically. Now you can put that numerical value for y back into x = 20 - y to get a numerical answer for x.

The same process works for bigger sytems. At each step you, you get an algebraic expression that equals one of the unknowns and substitute that into the remaining equations until you are down to a single equation in one unknown. That is all you need to know to solve any system of n linear equations in n unknowns. (In practice, when n is larger than a relatively small number, more efficient techniques are used,)

EDIT: While I was writing pka gave a slightly different way to solve the same system. Both methods will lead to the same result. His approach was chosen because it makes the mechanical part of the problem somewhat easier. (Can you tell us why?) As you get experience, you too will be able to see ways to attack a problem that will be easier mechanically. It is important to know that there can be multiple ways to arrive at a correct answer.
 
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Thanks

Hello again,

Thank you all for your help. You'll be pleased to know that as a result, I'm now solving these problems with ease. I have learned the elimination method and the substitution method (these two are both ways of manipulating an equation with two or more unknowns and thus transforming it into to a solvable equation containing only one unknown), I've also learned Cramer's rule, which is quite different from the first two methods. Each of these three techniques would seem to lend itself better to a different linear equation situation, and I guess this is where experience comes in: knowing which one to use in different situations. I'd also like to thank Jomo, for linking me to that Youtube channel which contains a wealth of helpful videos.

Many thanks and the warmest regards,

Tom.
 
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