Help me to solve this math problem? "Three foreign languages are offered at school."

bleuetrose

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Help me to solve this math problem? "Three foreign languages are offered at school."

May you please help me to solve this math task? It is very hard for me and I cannot help my child to understand how is going to the results. I know only the answer but how this solution come it is not clear for me.

This is the math task:
Three foreign languages are offered in Lincoln Elementary School: Spanish, French and Chinese. In Mrs. Moore's class, every student takes one and only one foreign language class. There are 14 students taking either Spanish or French. There are 18 students taking either Spanish or Chinese. And there are 16 students taking either French or Chinese. How many students in Mrs. Moore's class are taking Chinese?

Thank you.
 
… I cannot help my child to understand …
Can your child work with us directly? We like students to show us any work they have started, so we can see where they got stuck.


Three foreign languages are offered in Lincoln Elementary School: Spanish, French and Chinese. In Mrs. Moore's class, every student takes one and only one foreign language class. There are 14 students taking either Spanish or French. There are 18 students taking either Spanish or Chinese. And there are 16 students taking either French or Chinese. How many students in Mrs. Moore's class are taking Chinese?
In the first step, we assign different symbols to represent the unknown quantities. In the next step, we use those symbols to translate sentences into math equations.

Here's an example.

We have some apples, bananas, and cherries. If we put the apples and bananas into a box, there are 14 pieces of fruit. If we put the bananas and cherries into a box, there are 49 pieces of fruit. If we put the apples and cherries into a box, there are 45 pieces of fruit. How many cherries do we have?

The first step is to assign symbols (we call them "variables") to represent the unknown quantities.

Let A = the number of apples

Let B = the number of bananas

Let C = the number of cherries

The next step is to translate sentences into equations.

We're told that the apples and bananas together are 14 pieces of fruit.

A + B = 14

The bananas and cherries together total 49 pieces.

B + C = 49

The apples and cherries together total 45 pieces.

A + C = 45

The question asks, "How many cherries do we have?" We need to find the value of C.

Now we start solving an equation for one variable, and substituting the result into another equation. Then we repeat. The goal is to get an equation that contains only variable C.

There is more than one way to do this. If we are not sure where to begin, then we can just guess, and see what happens. With enough practice, we will not have to guess.

B + C = 49

Solve this equation for B

B + C - C = 49 - C

B = 49 - C

Now we have an expression for symbol B that we can substitute into another equation.

A + B = 14

A + 49 - C = 14

Solve this equation for A

A + 49 - C - 49 + C = 14 - 49 + C

A = -35 + C

Now we have an expression for A that contains only symbol C. We substitute it into the remaining equation.

A + C = 45

-35 + C + C = 45

We have achieved our goal; this equation contains only the variable C.

Solve the equation for C

-35 + 2C = 45

-35 + 2C + 35 = 45 + 35

2C = 80

C = 40

There are 40 cherries.

The same method can be used to find the number of students taking Chinese.

If your child cannot understand this example, make sure that they are not simply reading it. They need to copy every step on paper and work through the whole example. If your child gets stuck, then they can come back and explain to us which part(s) are confusing and why, so that we know where to continue helping. Thank you.
 
May you please help me to solve this math task? It is very hard for me and I cannot help my child to understand how is going to the results. I know only the answer but how this solution come it is not clear for me.

This is the math task:
Three foreign languages are offered in Lincoln Elementary School: Spanish, French and Chinese. In Mrs. Moore's class, every student takes one and only one foreign language class. There are 14 students taking either Spanish or French. There are 18 students taking either Spanish or Chinese. And there are 16 students taking either French or Chinese. How many students in Mrs. Moore's class are taking Chinese?

Thank you.
Let S be the number of students taking Spanish, F the number taking French, and C the number taking Chinese.

In Mrs. Moore's class, every student takes one and only one foreign language class.

There are 14 students taking either Spanish or French.
So S+ F= 14.

There are 18 students taking either Spanish or Chinese.
So S+ C= 18.

And there are 16 students taking either French or Chinese. How many students in Mrs. Moore's class are taking Chinese?
So F+ C= 16.
That gives three equations to solve for the three values S, F, and C. For example, S+ F= 14 and S+ C= 18 both involve "S" while F+ C= 16 does not. If we subtract S+ F= 14 from S+ C= 18 we get (S+ C)- (S+ F)= C- F= 18- 14= 4.

Now, we have C- F= 4 and C+ F= 16. What do you get if you add the those two equations?
 
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Thank you for your answers. Sorry, my child cannot write directly and it is complicated to explain to you why, but I may write what she did in order to find the solutions. Thank you for trying to help me.

HallsofIvy, we like the way that you show me a solution for this math tack. We followed you until here:
That gives three equations to solve for the three values S, F, and C. For example, S+ F= 14 and S+ C= 18 both involve "S" while F+ C= 16 does not. If we subtract S+ F= 14 from S+ C= 18 we get (S+ C)- (S+ F)= C- F= 18- 14= 4.

But after we got confused, what is difference 4 stand for? How this show how much is Spanish and French and Chinese? May you please explain this one more detail pls?

Yes, if we continue:
C- F= 4 and C+ F= 16 will be
(
C- F)+(C+ F)= 4+16
C-F+C+F = 4+16
2C=20
C=20/2
C=10 and this is the right answer. Thank you. Sorry, we still didn't understand this tack. May you be a little more detailed? In this way, we will understand more.

What this 4 stand for on the top? What that difference gave you about the quantity of the Spanish, French, and Chinese?

My child did that:
Every student takes one and only one foreign language class, so

The first guess:
7 (Spanish) + 7 (French) = 14
7(Spanish) + 11 (Chinese) = 18, and then it become a problem, because
7 (French) + 11 (Chinese) = 18 (should be 16)

So, she changed the guessed numbers started from the last line French + Chinese
The second guess:
9 (Spanish) + 7(French) = 14 became problem, because should be 14, not 16
9(Spanish) + 9 (Chinese) = 18, and then it become a problem, because
7 (French) + 9 (Chinese) = 18 (should be 16)

So when she continues to change the numbers, one of the lines going wrong.





 
And also why we add in the end these instead of subscribing the difference between them C- F= 4 and C+ F= 16.

Sorry, we confused.
HallsofIvy, in this way that you explained we understand, but what this difference 4 is between Chinese and French? And if that means that 4 students speak more Chinese than French, so what that help about the ratio between French, Spanish and Chinese? Pls for more details.

Thank you.
 
If we subtract S+ F= 14 from S+ C= 18 we get (S+ C)- (S+ F)= C- F= 18- 14= 4.

Now, we have C- F= 4 and C+ F= 16. What do you get if you add the those two equations?

And also why we add in the end these instead of subscribing the difference between them C- F= 4 and C+ F= 16.

Sorry, we confused.
HallsofIvy, in this way that you explained we understand, but what this difference 4 is between Chinese and French? And if that means that 4 students speak more Chinese than French, so what that help about the ratio between French, Spanish and Chinese? Pls for more details.

The point of Halls' work is not that the number 4 in itself is important, but that he obtained an equation involving only C and F. So in the end, he has two equations in two unknowns: C-F = 4 and C+F = 16. These can then be solved similar to what he already did: add the two equations to eliminate F: (C-F) + (C+F) = 4 + 16, which gives 2C = 20, from which we know that C = 10. From that, you can find the values of S and F.

All this will be useless unless you and your child both have learned how to solve a system of equations. If not, then we need to find another way.

What your child did suggests that she does not necessarily know algebra at all. It will help if you can tell us something about what she has learned. Has she seen any algebra at all? Has she seen any examples of these problems solved by anything other than "guess and check"?
 

My child did that:
Every student takes one and only one foreign language class, so

The first guess:
7 (Spanish) + 7 (French) = 14
7(Spanish) + 11 (Chinese) = 18, and then it become a problem, because
7 (French) + 11 (Chinese) = 18 (should be 16)

So, she changed the guessed numbers started from the last line French + Chinese
The second guess:
9 (Spanish) + 7(French) = 14 became problem, because should be 14, not 16
9(Spanish) + 9 (Chinese) = 18, and then it become a problem, because
7 (French) + 9 (Chinese) = 18 (should be 16)

So when she continues to change the numbers, one of the lines going wrong.

If I wanted to do this by guessing, I would guess just one number, find the results, and change that same one as needed. Doing something different each time is the way to discourage yourself. Following a consistent pattern leads to success.

The facts are:

S + F = 14
S + C = 18
F + C = 16

Since we want to find C, make a guess for that. We know it can't be greater than 16; so let's try 8.

Since S + C = 18, S = 18 - C. (That is, the number taking Spanish is 18 minus the number taking Chinese.)
Since F + C = 16, F = 16 - C. (That is, the number taking French is 16 minus the number taking Chinese.)

For our guess, S = 18 - 8 = 10; F = 16 - 8 = 8.

Then we check the other fact: S + F = 10 + 8 = 18. That's too many.

Now, can you see a reason to either increase or decrease our guess? You might; if you don't, then just choose one (that is, either increase C to 9 or decrease it to 7), and try again. If things have moved in the wrong direction, do the opposite of what you did. If it moved in the right direction, but is not solved yet, repeat.

There are more efficient ways to guess, but if guessing is the method of choice, then the way to become good at it is to keep trying, and after you succeed, look back to see if you could have done it more quickly. Discovering better ways by yourself is the best way to learn something like this.
 
If I wanted to do this by guessing, I would guess just one number, find the results, and change that same one as needed. Doing something different each time is the way to discourage yourself. Following a consistent pattern leads to success.

The facts are:
S + F = 14
S + C = 18
F + C = 16

Since we want to find C, make a guess for that. We know it can't be greater than 16; so let's try 8.

Since S + C = 18, S = 18 - C. (That is, the number taking Spanish is 18 minus the number taking Chinese.)
Since F + C = 16, F = 16 - C. (That is, the number taking French is 16 minus the number taking Chinese.)

For our guess, S = 18 - 8 = 10; F = 16 - 8 = 8.

Then we check the other fact: S + F = 10 + 8 = 18. That's too many.

Now, can you see a reason to either increase or decrease our guess? You might; if you don't, then just choose one (that is, either increase C to 9 or decrease it to 7), and try again. If things have moved in the wrong direction, do the opposite of what you did. If it moved in the right direction, but is not solved yet, repeat.

There are more efficient ways to guess, but if guessing is the method of choice, then the way to become good at it is to keep trying, and after you succeed, look back to see if you could have done it more quickly. Discovering better ways by yourself is the best way to learn something like this.

Thank you very much!
 
The point of Halls' work is not that the number 4 in itself is important, but that he obtained an equation involving only C and F. So in the end, he has two equations in two unknowns: C-F = 4 and C+F = 16. These can then be solved similar to what he already did: add the two equations to eliminate F: (C-F) + (C+F) = 4 + 16, which gives 2C = 20, from which we know that C = 10. From that, you can find the values of S and F.

All this will be useless unless you and your child both have learned how to solve a system of equations. If not, then we need to find another way.

What your child did suggests that she does not necessarily know algebra at all. It will help if you can tell us something about what she has learned. Has she seen any algebra at all? Has she seen any examples of these problems solved by anything other than "guess and check"?

Thank you very much!
 
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