Please help! I'm stuck!

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Starfire

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Ok i dont know if my equation makes sense or not as im touching algebra after a long time so Im facing problems with this simplification. Please help.

=> -2ab^2 + ab + 138b - 69 = 0

I got stuck here.
The equation started like this...

ab + ab = 138, find A. ?
 
Starfire said:
Ok i dont know if my equation makes sense or not as im touching algebra after a long time so Im facing problems with this simplification. Please help.

=> -2ab^2 + ab + 138b - 69 = 0

I got stuck here.
The equation started like this...

ab + ab = 138, find A. ?
Click to expand...
I do not see A in the equation

ab + ab = 138

And please show us how you arrived at:

-2ab^2 + ab + 138b - 69 = 0
 
Subhotosh Khan said:
I do not see A in the equation

ab + ab = 138

And please show us how you arrived at:

-2ab^2 + ab + 138b - 69 = 0
Click to expand...
Oh sorry, that was a typo. Its find 'a'.
And about how I arrived at that equation, I'm pretty sure I'm way off the mark so it'd be great help if the value of 'a' can be found using the original equation. If you could also share the whole process, it'd be great!
 
Starfire said:
Ok i dont know if my equation makes sense or not as im touching algebra after a long time so Im facing problems with this simplification. Please help.

=> -2ab^2 + ab + 138b - 69 = 0

I got stuck here.
The equation started like this...

ab + ab = 138, find A. ?
Click to expand...
So 2ab= 138, a= 138/2b= 69/b? Without more information, that's as far as you can go. I don't see how you could get the previous equation from that.
 
Starfire said:
Oh sorry, that was a typo. Its find 'a'.
And about how I arrived at that equation, I'm pretty sure I'm way off the mark so it'd be great help if the value of 'a' can be found using the original equation. If you could also share the whole process, it'd be great!
Click to expand...
Please repost the corrected problem along with your work - even if you know it is incorrect. That will give us a "starting point" of further discussion.
 
If you add something to itself, regardless of how unfriendly the something looks, you get 2 times that something.

5+5 = 2*5

log(7) + log(7) = 2log(7)

slice of pie + slice of pie = 2 slices of pie, ie 2*(slice of pie)

So surely ab + ab = 2ab.
 
Speaking a bit loosely, if you have n unknowns, you need n equations to find their numeric value.

Thus, all of us suspect that there may be information that you have not supplied. For example

[MATH]ab + ab = 138 \text { and } b - a = 10 \implies[/MATH]
[MATH]2ab = 138 \text { and } b = a + 10 \implies 69 = ab = a(a + 10) \implies[/MATH]
[MATH]a^2 + 10a - 69 = 0.[/MATH]
But we have no clue whether that is the missing information or something else.
 
Last edited:
HallsofIvy said:
So 2ab= 138, a= 138/2b= 69/b? Without more information, that's as far as you can go. I don't see how you could get the previous equation from that.
Click to expand...
i also think so..but the question had no other info other than that ?
 
JeffM said:
Speaking a bit loosely, if you have n unknowns, you need n equations to find their numeric value.

Thus, all of us suspect that there may be information that you have not supplied. For example

[MATH]ab + ab = 138 \text { and } b - a = 10 \implies[/MATH]
[MATH]2ab = 138 \text { and } b = a + 10 \implies 69 = ab = a(a + 10) \implies[/MATH]
[MATH]a^2 + 10a - 69 = 0.[/MATH]
But we have no clue whether that is the missing information or something else.
Click to expand...
i have the same suspicions.. i have posted how the original question looked like.. and also how i came to my weird equation..
 
AB (2).png
so this was how the question was actually given..
and below was how i managed to create a hotchpotch of an equation..?
IMG_20200929_111044.jpg

p.s. there was no other info given and im not sure if there's an answer but it'd be great if there was a way to find the answer with just the available question ..
 
If AB+ AB= 2AB= 138 then AB= 138/2= 69. So you can interpret that as saying that the two digit number AB is 69 or as saying that the product of the two numbers A and B is 69= 3(23). If A and B are any real numbers, there are infinitely many solutions. If A and B are positive integers, one of A and B is 3 and the other is 23.
 
There are puzzles where AB stands for a two digit number. People who enjoy such puzzles do not usually attack them through equations but through a set of simple recursive algorithms, but they can be solved by means of so-called Diophantine equations.

In this kind of puzzle, AB + AB = 138 means mathematically

[MATH]a \text { and } b \text { are whole numbers such that}\\ 1\le a \le 9,\ 0 \le b \le 9,\text { and } (10a + b) + (10 + b) = 138.[/MATH]We have two unknowns and only one equation. In my first post, I said this is not generally solvable unless there is a second equation. But here we have a lot of additional information that allows us to to generate useful inequations after we generate more unknowns

[MATH](10a + b) + (10a + b) = 138 \implies 138 = 20a + 2b = 10(2a) + 2b.[/MATH]
[MATH]\text {There exist non-negative integers } u,\ v,\ x, \text { and } y \text { such that} \\ 2a = 10u + v, \ 0 \le u \le 1,\ 0 \le v \le 8,\ v \text { is even,}\\ 2b = 10x + y, \ 0 \le x \le 1, \ 0 \le y \le 8, \text { and } y \text { is even.}[/MATH][MATH]\therefore 138 = 10(2a) + 2b = 10(10u + v) + 10x + y = 100u + 10(v + x) + y \implies \\ 100u + 10(v + x) = 138 - y.[/MATH]Note that u, v, and x are all integers. Therefore 10u + v + x is an integer.

[MATH]100u + 10(v + x) = 138 - y \text { and } 0 \le y \le 8 \implies \\ 130 \le 100u + 10(v + x) \le 138 \implies 13 \le 10u + v + x \le 13.8 \implies \\ 13 = 10u + v + x \text { and } y = 8.[/MATH]Note that [MATH]v + x \le 9.[/MATH]
We have two possible cases for u, u = 0 or u = 1.

[MATH]u = 0 \implies 13 = 10u + v + x = v + x \le 9 \implies \\ 13 \le 9, \text { which is nonsense} \\ u = 1 \implies 10(1) + v + x = 13 \implies 3 = v + x.[/MATH]We have two possible cases for x, x = 0 or x = 1.

[MATH]x = 0 \implies 3 = v + x \implies v = 3, \text { which is }\\ \text{impossible because } v \text { is even} \implies \\ x = 1 \implies v = 2.[/MATH][MATH]\therefore 2a = 10u + v = 10(1) + 2 = 12 \implies a = 6. \text {And } 2b = 10x + y = 10(1) + 8 = 18 \implies b = 9.[/MATH]
Thus AB is 69. So it can be solved through equations and inequations not that anyone sane would do it that way.

Here is how I would actually do it.

2 times some integer is 138. That means the integer's unit digit must be 4 or 9. It it is 4 then we have

2(10p + 4) = 138, which means 20p = 130, which means that p = 6.5, which is impossible.

Thus the unit's digit must be 9.

2(10p + 9) = 138, which means 20p = 138 - 18 = 120 so p = 6.
 
JeffM says
"In this kind of puzzle, AB + AB = 138 means mathematically
a and b are whole numbers such that1≤a≤9, 0≤b≤9, and (10a+b)+(10+b)=138.a and b are whole numbers such that1≤a≤9, 0≤b≤9, and (10a+b)+(10+b)=138."

Well maybe but the puzzle was presented as

AB (2).png


rather suggesting an adding by columns approach.
 
JayJay said:
JeffM says
"In this kind of puzzle, AB + AB = 138 means mathematically
a and b are whole numbers such that1≤a≤9, 0≤b≤9, and (10a+b)+(10+b)=138.a and b are whole numbers such that1≤a≤9, 0≤b≤9, and (10a+b)+(10+b)=138."

Well maybe but the puzzle was presented as

AB (2).png


rather suggesting an adding by columns approach.
Click to expand...
Yes, exactly. But the OP initially tried to solve it algebraically. I always think it is sensible to show someone that their intuition was not wrong. They can do it that way. Then show them that there is an easier way.

It is not, in my opinion, a good idea to discourage native intuition. Others are free to disagree.
 
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