Help please, I'm struggling to figure out a simpler way of working this out for a 13 year old

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Raanikeri

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The question is :
1652119850617.png

This was a question taken from a resource meant for 10 year olds. I'm struggling to find an easier way of solving this than going into loads of algebraic working involving algebraic factorising. I'm sure it doesn't require factorisation as it's meant for 10 year olds, but I have no idea how to work it out in a simpler way.
Any help please?
Thanks in advance!
 
Raanikeri said:
The question is :
View attachment 32658

This was a question taken from a resource meant for 10 year olds. I'm struggling to find an easier way of solving this than going into loads of algebraic working involving algebraic factorising. I'm sure it doesn't require factorisation as it's meant for 10 year olds, but I have no idea how to work it out in a simpler way.
Any help please?
Thanks in advance!
Click to expand...
It's probably intended to be solved by trial and error. At worst, just try A=1, B=12, and so on. Or, start with A and B as close as possible ...

Alternatively, since AB=3*14, you could try different factorizations of 42, looking for a sum of 13. Oddly enough, this is the same trial and error technique used to factor a quadratic.
 
Dr.Peterson said:
It's probably intended to be solved by trial and error. At worst, just try A=1, B=12, and so on. Or, start with A and B as close as possible ...

Alternatively, since AB=3*14, you could try different factorizations of 42, looking for a sum of 13. Oddly enough, this is the same trial and error technique used to factor a quadratic.
Click to expand...
Thank you! That makes sense!

Yes I did the factorising and got B=6, A=7..
 
Raanikeri said:
Thank you! That makes sense!

Yes I did the factorising and got B=6, A=7..
Click to expand...
There are actually 2 solutions for C since addition and multiplication are commutative, i.e. \(\displaystyle AB=BA\) and \(\displaystyle A+B=B+A\).
WIth that said, the solution pair (6,7) vs. (7,6) will yield different results for C.
 
BigBeachBanana said:
There are actually 2 solutions for C since addition and multiplication are commutative, i.e. \(\displaystyle AB=BA\) and \(\displaystyle A+B=B+A\).
WIth that said, the solution pair (6,7) vs. (7,6) will yield different results for C.
Click to expand...
Yeah I got 2 solutions, but only B=6 and A=7 will work in place of the missing digits in the equivalent fractions shown in the question. If B=7, and A=6, then the fractions shown will not be equivalent.
 
Raanikeri said:
Yeah I got 2 solutions, but only B=6 and A=7 will work in place of the missing digits in the equivalent fractions shown in the question. If B=7, and A=6, then the fractions shown will not be equivalent.
Click to expand...
I would have to disagree.
For A=6, B=7, will give C=24.
[math]\frac{3}{6}=\frac{7}{14}=\frac{12}{\red{24}}=\frac{1}{2}[/math]
 
Here is a simple (but mathematically inconsistent) solution:-

\(\displaystyle \frac{3}{A}=\frac{B}{14} \Rightarrow \frac{3×14}{14×A}=\frac{A×B}{14×A} \Rightarrow A×B=42\)       (as Dr.P. points out)

If A×B=42 & A+B=13 that suggests* that A=6 & B=7 (or vice versa)
and 'trying' A=6 & B=7 that gives us a nice "solution" where
(all) the fractions are now equivalent to \(\displaystyle \frac{1}{2}\) and thence, C=24.

*What two numbers add up to 13 and multiply to 42?
​
This method of "working it out" is mathematically "inconsistent" because the derived result: A×B=42 combined with A+B=13 leads us (mathematicians) inexorably to \(\displaystyle A^2-13A-42=0 \Rightarrow A \approx\) 15.68 or -2.68 which results lead to some of the other possible values for C that BigBeachBanana refers to.
NB: I started 'composing' this Reply before I had seen all the other posts that were submitted after DrPetersen's; I probably wouldn't have bothered starting it if I had seen them but, as they say in Mastermind: I've started so.... ?

Clearly, however, your pupils don't need to know any of that so you just show them the first two lines of the "working" (assuming your "13 year old" is familiar with cross-multiplication of fractions) and tell them to "figure out" whether it's A=6 & B=7 or vice versa. ?
 
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The Highlander said:
\(\displaystyle A^2-13A\red{−}42=0\)
Click to expand...
It should be a plus before 42, [imath]A^2-13A+42=(A-6)(A-7)=0[/imath] which give 2 integer solutions [imath]A=6,7[/imath] as expected.
My point was for the pair (A,B) will yield different result of C compared to (B,A), i.e. (6,7) -> C=24, whereas (7,6) -> C=28.
 
BigBeachBanana said:
It should be a plus before 42, [imath]A^2-13A+42=(A-6)(A-7)=0[/imath] which give 2 integer solutions [imath]A=6,7[/imath] as expected.
My point was for the pair (A,B) will yield different result of C compared to (B,A), i.e. (6,7) -> C=24, whereas (7,6) -> C=28.
Click to expand...
You're right, of course, I stupidly set B=(A-13) instead of B=(13-A) which gave me [imath]A^2-13A-42=0[/imath] and thence the irrational roots! ?
I was sorely disappointed when the quadratic formulated from the initial expressions didn't give me the integer roots I was expecting* and despite going over my algebraic manipulation repeatedly (in the hope of finding some way to get the (desired) results of 6 & 7) I failed to spot my initial mistake! Doh! ?

*Because 6 & 7 were such 'convenient' factors of 42 that added to 13 and I could already see that A=6 & B=7 fitted the question giving C=24! ?
 
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