MHB Simple Algebra? This is from a 10 year olds homework

[Rob Murphy]

OK, either I'm looking at this the wrong way, or this is way above what a 10 year old should be.

Following are 2 equations containing A, B and C. A solution must be found that solves both equations, i.e. A, B and C are the same in each formula. Also, A, B and C must be whole numbers;

C+2A=3B+2

3C+4A=8B

Can anyone find values for A, B and C which solve both equations?

Thanks

Rob

 

[MarkFL]

I would begin by solving both equations for $4A$ to get a relationship between $B$ and $C$...what do you find?

 

[Jameson]

Hi Rob Murphy,

Welcome to MHB! :)

I'll also add that I'm not sure how a teacher would want a 10 year old to go about solving this. With 3 variables (A,B,C) and 2 equations we won't get one answer. There will be an infinite amount of answers. My guess would be they want the children to "guess and check" at this age, but we can find an example solution much faster like Mark is suggesting.

 

[Rob Murphy]

Hi Rob Murphy,

Welcome to MHB! :)

I'll also add that I'm not sure how a teacher would want a 10 year old to go about solving this. With 3 variables (A,B,C) and 2 equations we won't get one answer. There will be an infinite amount of answers. My guess would be they want the children to "guess and check" at this age, but we can find an example solution much faster like Mark is suggesting.

Thanks guys, I'm glad I hadn't missed anything obvious. I guess the question was not explained enough in that both equations probably don't need to the same values for A, B and C...

If the equations are taken in isolation, it's much easier for a 10 year old to solve by trial and error (posh for guessing...)

However, I'm fairly decent at maths, but I wouldn't know how to approach this in the original context I placed on the question, i.e. how can common values for A, B and C be calculated to solve both equations...

You say there would be an infinite amount of answers?

Could you help me calculate one?

Thanks

 

[Jameson]

Yes, there are an infinite amount of answers because there are 3 unknowns but only two restrictions. That leaves a free variable, as we sometimes call it. Again, this is WAY above what a 10 year old would be doing or at least not how they would be hearing the material.

To solve this problem in a general fashion, what you would do is solve the equations in terms of a free variable. That means if you have $A$ as your free variable, then $B$ would equal something in terms of $A$ and $C$ would equal something in terms of $A$ instead of static numbers. If you plug these equations into Wolfram you'll see that the solutions are in this form.

However this is too much for a 10 year old I think.

What I would do is something like this. Pick $A=1$ and our two equations simplify down to: $C+2=3B+2$ and $3C+4=8B$. Any idea how to solve for $B$ and $C$ now?

 

[Rob Murphy]

Yes, there are an infinite amount of answers because there are 3 unknowns but only two restrictions. That leaves a free variable, as we sometimes call it. Again, this is WAY above what a 10 year old would be doing or at least not how they would be hearing the material.

To solve this problem in a general fashion, what you would do is solve the equations in terms of a free variable. That means if you have $A$ as your free variable, then $B$ would equal something in terms of $A$ and $C$ would equal something in terms of $A$ instead of static numbers. If you plug these equations into Wolfram you'll see that the solutions are in this form.

However this is too much for a 10 year old I think.

What I would do is something like this. Pick $A=1$ and our two equations simplify down to: $C+2=3B+2$ and $3C+4=8B$. Any idea how to solve for $B$ and $C$ now?

Hi Jameson

I see the breakdown on Wolfram, I'm afraid that's a bit too much for me. Been a while since I did complex algebra.

But can I please ask, can the B=2A-6, C=4A-16 solution be translated into actual, single whole digits; i.e.

A=6
B=8
C=10

(obviously these are incorrect)

I'd just like to be able to see one actual answer.

Thanks

Rob

 

[Jameson]

Yes, I can give you an example answer but without showing how you got there it might be bad. Maybe the teacher just wants an example answer though.

Wolfram Alpha shows that if we take $A$ to be our free variable, then we can solve for $B$ and $C$ and get $B=2A-6$ and $C=4A-16$. So we let $A=1$, then it follows that $B=-4$ and $C=-12$. How can we check that this is a valid solution?

 

[Rob Murphy]

Yes, I can give you an example answer but without showing how you got there it might be bad. Maybe the teacher just wants an example answer though.

Wolfram Alpha shows that if we take $A$ to be our free variable, then we can solve for $B$ and $C$ and get $B=2A-6$ and $C=4A-16$. So we let $A=1$, then it follows that $B=-4$ and $C=-12$. How can we check that this is a valid solution?

That is superb

C+2A=3B+2
-12+2=-12+2
-10=-10

3C+4A=8B
-36+4=-32
-32=-32

Tremendous

Many thanks

Rob

 

[MarkFL]

Here's what I did:

The first equation implies:

$$4A=6B+4-2C$$

And the second equation implies:

$$4A=8B-3C$$

Now equating the two expressions for $4A$, we obtain:

$$C=2(B-2)$$

Plugging this into the second equation, we then find:

$$A=\frac{B}{2}+3$$

Thus, for $n\in\mathbb{N}$ (since all solutions must be whole numbers, or non-negative integers), we obtain:

$$(A,B,C)=\left(n+3,2n,4(n-1)\right)$$