A Very Nice Algebra Problem | Fill in the boxes

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Күн бұрын

Пікірлер: 33

@davidellis1929 3 ай бұрын

You can solve the problem with a single variable x representing the lower left box. The upper left box equals 8-x and the lower right box equals x-5. Then the upper right box can be expressed as both 21/(8-x) and 9/(x-5). Equating these two values yields x = 5.9, with the other three boxes equal to 2.1, 0.9 and 10.

@rosamariavercelloni1719 2 ай бұрын

Brilhante !

@adgf1x 17 күн бұрын

7×3=21 and 8-3=5

@greninjamariokartpokemonfan 2 ай бұрын

A×B=21, C-D=5, A+C=8, B×D=9 (A+C)-(C-D)=8-5 A+C-C+D=3 A+D=3 A=21/B, D=9/B 30/B=3 3B=30 B=10, A=2.1, D=0.9, C=5.9

@josefsezbrisko9064 2 ай бұрын

English : It is simply a system of 4 equations with 4 unknowns. There are several methods to solve such a system. The simplest is the elimination method. And perhaps the simplest solution would be to use a spreadsheet calculator. There, you only enter the constants and the solution is in a few seconds. Czech : Je to prostě soustava 4 rovnic o čtyřech neznámých. Metod jak takovou soustavu řešit je několik. Nejjednodušší je vylučovací metoda. A snad úplně nejjednodušeji by to šlo řešit pomocí tabulkového kalkulátoru. Tam se pouze zadají konstanty a řešení je v několika vteřinách.

@godot-lee 11 күн бұрын

B=10 A=21/10 C=59/10 D=9/10

@عبدالواسع-س8م 2 ай бұрын

Excellent. Thanks so much.

@freddyalvaradamaranon304 2 ай бұрын

Muchas gracias por compartir tan interesante video 😊😊. Realmente un sistema de ecuaciones con 4 incógnitas muy bien desarrollado y explicado.❤😊❤

@MrJacks13 26 күн бұрын

How do you derive to calculating (iii)-(ii)? Why not e.g. (i)+(iv)?

@lironne882 24 күн бұрын

що, не побачили в кінці помилку?

@gapalerg8044 11 күн бұрын

2,1 х10=21, 3х7=21 !!! 8-2,1=5,9. 10/9=0.9

@abbasmasum1633 3 ай бұрын

"d" must be negative. I used d as negative and it worked, but different result and worked.

@YardenVokerol 2 ай бұрын

a ↓ 8-a → 3-a ↑ 9/(3-a) a → 21/a 21/a=9/(3-a), 7(3-a)=3a, 21=10a, a=2.1 (2.1 10) (5.9 0.9)

@МишаВасинев 3 ай бұрын

Incredible!

@DanieleBarettin 3 ай бұрын

Nice presentation. But you need to be more specific: integer, rational...

@すぴか-z1q Ай бұрын

teacher! The number “9” written by the teacher only looks like a lowercase “g”!

@onuncukoy3404 Ай бұрын

empty works..

@andryvokubadra2644 3 ай бұрын

21 = 1*21, 3*7, 7*3, 21*1 9 = 1*9, 3*3, 9*1 But no one equal x + y = 8 or x - z = 5 So the roots ≠ Z 😅😅😅

@jamesharmon4994 3 ай бұрын

Nothing said it had to be integers.

@andryvokubadra2644 3 ай бұрын

@@jamesharmon4994 yes, But I proof that the root(s) aren't intenger 😊😊😊

@jamesharmon4994 3 ай бұрын

@@andryvokubadra2644 oh, ok. I knew that before watching the video. LOL

@ゆみ-p8b4u 2 ай бұрын

very easy.

@narine8044 17 күн бұрын

Thanks for reminding us But algebra and fragmentation are for the scholar Muhammad Musa Khwarazmi

@TBUGRATUIT-dm2vc 3 ай бұрын

+a×b-c=43 4+8-6+4=10 -d+e×f=15 1×4×8=32 a-8×d=-18 b+e=14 c+f=2

@nicatare2257 2 ай бұрын

3×3=9 9

@abba5512 22 күн бұрын

So fun

@olympiadeduworks 2 ай бұрын

great

@AbdelaKadir 2 ай бұрын

Very simple every body

@yanalkochaha2247 12 күн бұрын

why do you make it hard ? is it an easy way ? 7 x 3 = 21 then 1- 6 = 5 then from left up to down 7 +1 = 8 then from right up to down 3 +6 = 9 . be smart .

@FieriaAreilielle 5 күн бұрын

Retarded bot, glad that AI invasion will be far in the future..