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You can intuitively see that A and D are identical from A+B = 7 and D+B = 7. From that you can see that C+A = 10 and C-A= 1 gives you A=D=4.5 and C=5.5
@@SirWolf2018 I don't think that's the "most beautiful solution" because to say "intuitively see", it's not a proof anymore. By carrying out the proof of that and also the rest, it would have been just as long and time consuming as the way she did it. Also, her approach would work no matter what the numbers were. The A=D only works in select cases.
@@kenmore01 OK let me explain. In Math, trial and error can be just as practical sometimes as systematic approaches. Both are valid. When teaching, we can explain the systematic approach, but first you should keep the motivation high by explaining the most elegant solution first. Then we can talk about how the most elegant solution is applicable in general to every other similar puzzle. To me it's more important that Math doesn't feel like a chore, but you are also motivated to learn.
Κυρία μου, είστε πολύ συμπαθητική και τα προβληματάκια που βάζετε είναι ότι πρέπει για να "ακονίζεται" το μυαλό μου, τώρα που είμαι συνταξιούχος καθηγήτρια. Ήμουν φυσικός αλλά και τα μαθηματικά ήταν πάντα μεγάλη αγάπη μου. Μέχρι τώρα έχω λύσει όλα όσα είδα, και απλά επαληθεύω μετά το αποτέλεσμα, πράγμα που σημαίνει ότι μάλλον η άνοια ή το Αλτχάιμερ είναι μακριά ακόμη, χαχα! Σας ευχαριστώ πολύ, και σας χαιρετώ από την Ελλάδα. [Προτιμώ να γράφω στα ελληνικά, ξέροντας και ότι η πλατφόρμα δίνει μεταφράσεις].
a + b = 7 a + c = 10 b + d = 7 c - d = 1 a + b + a + c + b + d + c - d = 25, d's cancel so... 2a + 2b + 2c = 25 as a + b = 7, 2a + 2b = 14 subtract from above leaves... 2c = 11 c = 5.5
I used a 4 x 5 coefficient matrix - the first 4 columns represented the coefficients of A, B, C and D (all of them equal to 1) for each equation, and the 5th column represented the results of the 4 equations. Then by using standard row operations to reduce the number of coefficients, I obtained A=9/2, B=5/2, C=11/2, D=9/2. It may be a slightly longer process but it meant I didn't need to decide which equations to add/subtract. (I've recently been familiarizing myself with matrices and decided to do some practice). Love the channel by the way.
So I did this a bit differently. Since both A + B and B + D = 7, I made the assumption that A = D. Since A = D you can substitute D with A. Eventually you get C = 10 - A and C = 1 + A. Then 10 - A = 1 + A and A eventually becomes A = 9/2 or 4.5. Been a long time since I've done real math. Your videos are a fun reminding of what I loved about school.
I was thrown a bit by this as usually when these sorts of puzzles come up, the answers are integers. As there are only certain combinations of integers that add to 7 (7 and 0, 6 and 1, 5 and 2, 3 and 4) I tried all these and none of them worked. So then I tried the simultaneous equation method and when I got C=5.5 I thought I'd got it wrong!
if a+b = b+d ==> a = d and if a+c = 10 with also c-d = 1 and therefore c-a = 1 ==> 2a = 9 dc a =4.5 like d and after even un IQ of 80 can end ^^ a=4.5 d=4.5 b=1 c=5.5 30 sec top chrono
I used a+c = 10 and a+1 = c (a and d are the same since they both equal 7 when adding b) So, in a+c = 10 I substituted the c for a+1, which is a+a+1 = 10 Then, 2a = 9, which is a= 4.5 The rest was like you did with c = 5.5 but with a = 4.5, but as a bonus, I had d for free (also a 4.5).
Solution: You can immediately see, that A = D, because A + B = 7 and D + B = 7. So C + A = 10, but C - A = 1. Add both together to get 2C = 11, therefore C = 5.5. This gives us A = D = 10 - 5.5 = 4.5. Last but not least, we get B = 7 - 4.5 = 2.5. Done.
Can we not add 3rd & 4th equation to have B+C value. then add the 1st & 2nd equation and put the value of B+C in that to get the value of A B+D+C-D=7=1>B+C=8, 2A+B+C=17>2A+8=17>A=4.5
My immediate reaction was that if the bottom line was C+D it would have to total 10, since the total of the sums across must equal total of the sums downward. If we know that C+D=10 and C-D=1 then it is obvious that C=5.5 and D=4.5. Working out A and B takes another second or so, overall less than 10 seconds to get the 4 values. Writing out as many simultaneous equations as in the video seems massively longwinded.
Def started with since a+b=7 and b+d=7, therefore a=d. Then I did similar things for the rest, but getting rid of one variable immediately made it a bit easier.
for a in range(-20, 21): for b in range(-20, 21): for c in range(-20, 21): for d in range(-20,21): if a/2 + b/2 == 7: if a/2 + c/2 == 10: if c/2 - d/2 == 1: if b/2 + d/2 == 7: print (a/2,b/2,c/2,d/2) print("Hello, World!") 4.5 2.5 5.5 4.5 Hello, World!
kind of fun stuff, interetsing how everyone attacked this one. I solved for C by working the different equations down to answer that C was 5.5. Once you have that the rest fall out on the table. Cheers 😎
I needed pencil and paper for this one. With some simple algebra, solving the equations using pairs, the results are A = 9/2, B = 5/2, C = 11/2, and D = 9/2. I enjoy solving your “puzzles” before watching your videos.
We know that C = D+1 and therefore A = B+2 because the difference of 10 and 7 is 3. If A is 2 higher than B then the only way to resolve for 7 is A = 4.5 and B = 2.5.
I first noticed from the 1st and 3rd equations that a=d, both =7-b. She subtracts the new equations and gets -2c, adding them gets you 2b=5, (then b=2.5.
When there was just two equations involving B and C I did the subtraction the other way around. I subtracted the top equation from the bottom. That yields (B + C) - (B - C) = 8 - -3 which simplifies to 2C = 11. Basically the same as in the video without all those pesky negative signs to deal with at the end.
Didn't do it as nice and rigorous, but the easy start (to me at least) is in noticing that a+b=7 and b+d=7 implies that a must be equal to d. Then because the difference between c and a is only one, while added up together they add to 10 implies that 2a must equal 9. And thus a = 4.5, c = a + 1 = 5.5, b = 7 - a = 2.5 and d = a = 4.5.
From the second row we can see that D is one less than C. So if we replace the C in the first column with D, we get A+D=9 and since B+D=7 then A is 2 more than B So swapping the B for an A in the first row gives us A+A=9 So B=2.5, C=5.5, D=4.5 This was just the way I did it in my head from first glance but there are SO many ways to solve this.
You can solve this with two equations, not four: 1. A + B + C + D = 10 + 7 (adding the bottom sums) 2. A + B + C + D = 17 3. A + B + C - D = 7 + 1 (adding the right sums) 4. A + B + C - D = 8 Do Equation 2 - Equation 4. The A, B, and C drop out, and D - -D = 2D. 5. 2D = 9 6. D = 4.5 Once you know one, finding the others are equal.
I watched like 3-4 videos, it's seems like she only chooses easier questions, and she is little slow at solving, as in comments people are giving optimum (one liner) solutions. Please solve faster, and scale up the difficulty level.
Hey math friends! If you’re enjoying this video, could you double-check that you’ve liked it and subscribed to the channel? It’s a simple equation: your support + my passion = more great content! Thanks for helping me keep this going - you’re the best!
You can intuitively see that A and D are identical from A+B = 7 and D+B = 7. From that you can see that C+A = 10 and C-A= 1 gives you A=D=4.5 and C=5.5
by analysis, actually, but correct.
That's how I solved it too.
The fact that she didn't present the most beautiful solution must result in a video dislike. Sorry, but Math is Art if done right.
@@SirWolf2018 I don't think that's the "most beautiful solution" because to say "intuitively see", it's not a proof anymore. By carrying out the proof of that and also the rest, it would have been just as long and time consuming as the way she did it. Also, her approach would work no matter what the numbers were. The A=D only works in select cases.
@@kenmore01 OK let me explain. In Math, trial and error can be just as practical sometimes as systematic approaches. Both are valid. When teaching, we can explain the systematic approach, but first you should keep the motivation high by explaining the most elegant solution first. Then we can talk about how the most elegant solution is applicable in general to every other similar puzzle. To me it's more important that Math doesn't feel like a chore, but you are also motivated to learn.
Your voice is so soothing. You make maths beautiful again. Thank you.
I spent a minute trying to come up with a solution with only integers. It wasn't there so I just enjoyed the show.
Κυρία μου, είστε πολύ συμπαθητική και τα προβληματάκια που βάζετε είναι ότι πρέπει για να "ακονίζεται" το μυαλό μου, τώρα που είμαι συνταξιούχος καθηγήτρια. Ήμουν φυσικός αλλά και τα μαθηματικά ήταν πάντα μεγάλη αγάπη μου. Μέχρι τώρα έχω λύσει όλα όσα είδα, και απλά επαληθεύω μετά το αποτέλεσμα, πράγμα που σημαίνει ότι μάλλον η άνοια ή το Αλτχάιμερ είναι μακριά ακόμη, χαχα! Σας ευχαριστώ πολύ, και σας χαιρετώ από την Ελλάδα. [Προτιμώ να γράφω στα ελληνικά, ξέροντας και ότι η πλατφόρμα δίνει μεταφράσεις].
a + b = 7
a + c = 10
b + d = 7
c - d = 1
a + b + a + c + b + d + c - d = 25, d's cancel so...
2a + 2b + 2c = 25
as a + b = 7, 2a + 2b = 14 subtract from above leaves...
2c = 11
c = 5.5
I used a 4 x 5 coefficient matrix - the first 4 columns represented the coefficients of A, B, C and D (all of them equal to 1) for each equation, and the 5th column represented the results of the 4 equations. Then by using standard row operations to reduce the number of coefficients, I obtained A=9/2, B=5/2, C=11/2, D=9/2. It may be a slightly longer process but it meant I didn't need to decide which equations to add/subtract. (I've recently been familiarizing myself with matrices and decided to do some practice). Love the channel by the way.
So I did this a bit differently. Since both A + B and B + D = 7, I made the assumption that A = D. Since A = D you can substitute D with A. Eventually you get C = 10 - A and C = 1 + A. Then 10 - A = 1 + A and A eventually becomes A = 9/2 or 4.5. Been a long time since I've done real math. Your videos are a fun reminding of what I loved about school.
I learn scientific English by listening your lessons.Thank you cute teacher.
if A+B = 7 and B+D = 7, A=D.
so, A+C = 10 and C-A* = 1
A = 10-C
C-(10-C) = 1
C -10 + C = 1
2C = 11
C = 5.5
A = D = 4,5
B = 2,5
I first thought that all numbers had to be integers, but none of the numbers I tried worked. 😅 So I watched your video. Thank you!
I solved with sympy: "import sympy as sp"; "A,B,C,D = sp.symbols('A,B,C,D')"; "eqs = []"; "eqs.append(sp.Eq(A+B,7))";"eqs.append(sp.Eq(C-D,1))";"eqs.append(sp.Eq(A+C,10))";"eqs.append(sp.Eq(B+D,7))";"sp.solve(eqs)" results in {A: 9/2, B: 5/2, C: 11/2, D: 9/2}.
Excellent teaching style as always!
I was thrown a bit by this as usually when these sorts of puzzles come up, the answers are integers.
As there are only certain combinations of integers that add to 7 (7 and 0, 6 and 1, 5 and 2, 3 and 4) I tried all these and none of them worked. So then I tried the simultaneous equation method and when I got C=5.5 I thought I'd got it wrong!
Thank you very much and much success for your new Channel!! 💙🐬
Woow, this is nice
Great, that you liked it!
How I solved it:
Add equations 1-4:
(2*A)+(2*b)+(2*c) = 25
a+b+c = 12.5
from eq. 1: c=5.5
from eq. 2: a=4.5
from eq. 1: b=2.5
from eq. 4: d=4.5
if a+b = b+d ==> a = d
and if a+c = 10 with also c-d = 1 and therefore c-a = 1 ==> 2a = 9 dc a =4.5 like d
and after even un IQ of 80 can end ^^
a=4.5
d=4.5
b=1
c=5.5
30 sec top chrono
Great teaching
Thank you so much ❤❤
From the columns we can see that A+B+C+D = 17
From the rows we can see that A+B+C-D = 8
So 2D = 9, D = 4½. So C = 5½, A = 4½, B = 2½.
I used a+c = 10 and a+1 = c (a and d are the same since they both equal 7 when adding b)
So, in a+c = 10 I substituted the c for a+1, which is a+a+1 = 10
Then, 2a = 9, which is a= 4.5
The rest was like you did with c = 5.5 but with a = 4.5, but as a bonus, I had d for free (also a 4.5).
A nice practice of RREF on my good old TI-92 Plus. Reduced Row Echelon Form
awesome mam
3:36 Just add the two equations and eliminate C, it's much easier. you get 2B = 5.
Solution:
You can immediately see, that A = D, because A + B = 7 and D + B = 7.
So C + A = 10, but C - A = 1. Add both together to get 2C = 11, therefore C = 5.5.
This gives us A = D = 10 - 5.5 = 4.5.
Last but not least, we get B = 7 - 4.5 = 2.5.
Done.
Can we not add 3rd & 4th equation to have B+C value. then add the 1st & 2nd equation and put the value of B+C in that to get the value of A
B+D+C-D=7=1>B+C=8,
2A+B+C=17>2A+8=17>A=4.5
My immediate reaction was that if the bottom line was C+D it would have to total 10, since the total of the sums across must equal total of the sums downward.
If we know that C+D=10 and C-D=1 then it is obvious that C=5.5 and D=4.5. Working out A and B takes another second or so, overall less than 10 seconds to get the 4 values. Writing out as many simultaneous equations as in the video seems massively longwinded.
Def started with since a+b=7 and b+d=7, therefore a=d.
Then I did similar things for the rest, but getting rid of one variable immediately made it a bit easier.
for a in range(-20, 21):
for b in range(-20, 21):
for c in range(-20, 21):
for d in range(-20,21):
if a/2 + b/2 == 7:
if a/2 + c/2 == 10:
if c/2 - d/2 == 1:
if b/2 + d/2 == 7:
print (a/2,b/2,c/2,d/2)
print("Hello, World!")
4.5 2.5 5.5 4.5
Hello, World!
I started with A+B=B+D => A=D and then substituted that back in.
A=4.5 B=2.5 C=5.5 D=4.5 final answer
kind of fun stuff, interetsing how everyone attacked this one. I solved for C by working the different equations down to answer that C was 5.5. Once you have that the rest fall out on the table. Cheers 😎
i did it with U
Awesome! Could you follow? 😊
@ yes
A=D=4.5
C=5.5
B=2.5
My way of solution ▶
A+B= 7
A+C= 10
B+D= 7
C-D= 1
⇒
A+B= 7
-A-C= -10
⇒
B-C= -3
C-B= 3
B+D= 7
C-B= 3
⇒
D+C= 10
C-D= 1
⇒
2C= 11
C= 11/2
C= 5,5
D+C= 10
C= 5,5
⇒
D= 4,5
B+D= 7
D= 4,5
⇒
B= 2,5
A+B= 7
B=2,5
⇒
A= 4,5
𝕃 = { 4,5 ; 2,5 ; 5,5 ; 4,5 }
Quite easy to see that A = D so there are 4 equations and 3 unknown and therefor really too easy to solve.
Better make it a 4 by 4.
I needed pencil and paper for this one. With some simple algebra, solving the equations using pairs, the results are A = 9/2, B = 5/2, C = 11/2, and D = 9/2. I enjoy solving your “puzzles” before watching your videos.
nice
Immediately knew that A=D because B was in 2 problems where the answer was the same in this case 7.
We know that C = D+1 and therefore A = B+2 because the difference of 10 and 7 is 3. If A is 2 higher than B then the only way to resolve for 7 is A = 4.5 and B = 2.5.
(10 +7) - (7 +1) = 2D
9 = 2D
D = 4.5
I immediately saw that A & D had to be equal.
Calculus please.
I gave up but I knew A = D lol
But you tried, which is great!
A bit deceiving from the start, because we are led to believe that a, b, c, and d are whole numbers. Maybe a disclaimer would help
Oh, You ended up with decimal fractions. I had expected just integers! (sorry, just a bad joke)
I first noticed from the 1st and 3rd equations that a=d, both =7-b.
She subtracts the new equations and gets -2c, adding them gets you 2b=5, (then b=2.5.
When there was just two equations involving B and C I did the subtraction the other way around. I subtracted the top equation from the bottom. That yields (B + C) - (B - C) = 8 - -3 which simplifies to 2C = 11. Basically the same as in the video without all those pesky negative signs to deal with at the end.
Question is wrong. If A:4.5 and D:4.5 you can’t say A and D. You can say A or D both of them.
Ich verstehe deine Englisch, aber warum kein Deutsch? UND BITTE, KEIN MEHR AI "Stimme/Sprache"! in different language!
Das hier ist mein neuer englischer Kanal. Auf meinem MathemaTrick Kanal gibt es diese Videos alle auch auf Deutsch 😊
Didn't do it as nice and rigorous, but the easy start (to me at least) is in noticing that a+b=7 and b+d=7 implies that a must be equal to d. Then because the difference between c and a is only one, while added up together they add to 10 implies that 2a must equal 9. And thus a = 4.5, c = a + 1 = 5.5, b = 7 - a = 2.5 and d = a = 4.5.
From the second row we can see that D is one less than C.
So if we replace the C in the first column with D, we get A+D=9
and since B+D=7 then A is 2 more than B
So swapping the B for an A in the first row gives us A+A=9
So B=2.5, C=5.5, D=4.5
This was just the way I did it in my head from first glance but there are SO many ways to solve this.
A+B= 7
A+C= 10
B+D= 7
C-D= 1
Solving further we get
B+C= 8
A+B = 7
A-C= -1
A+C= 10
2A= 9
A= 4.5, B= 2.5, C= 5.5, D= 4.5
Method 2
A+B= B+D
A=D
C+D= 10
C-D= 1
2C= 11
A= 4.5
B= 2.5
C= 5.5
D= 4.5
You are just doing re runs of another channel. 2 posts in the last 5 days have already been done by him.🤨
Du bist süß
You can solve this with two equations, not four:
1. A + B + C + D = 10 + 7 (adding the bottom sums)
2. A + B + C + D = 17
3. A + B + C - D = 7 + 1 (adding the right sums)
4. A + B + C - D = 8
Do Equation 2 - Equation 4. The A, B, and C drop out, and D - -D = 2D.
5. 2D = 9
6. D = 4.5
Once you know one, finding the others are equal.
Let’s do it in our head rather than algebra ok baby?
I watched like 3-4 videos, it's seems like she only chooses easier questions, and she is little slow at solving, as in comments people are giving optimum (one liner) solutions.
Please solve faster, and scale up the difficulty level.