Fill the Boxes | Justify your answer | (Puzzle) |

[@landesnorm]

That was fun!

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@marcgriselhubert3915]

a+b = 9; a.c = 36, b-d = 11, c+d = 7 d = 7-c, so: a+b = 9, a.c = 36, b+c = 18 d = 7-c, c = 18-b, so a+b =9, a.(18-b) =36 or 18.a-a.b = 36 d= 7-c, c= 18-b, b=9-a, so 18.a -a.(9-a) = 36 or a^2 +9.a -36 = 0 The second degree equation gives a = 3 or a = -12 Then (a,b,c,d) = (3,6,12,-5) or (-12,21,-3,10). No difficulty, just eliminate unknown letters step by step.

[@abeonthehill166]

Your step by step demonstration’s are so easy to follow . Thanks for sharing Man !

[@yalchingedikgedik8007]

That’s very nice and enjoyable Thanks Sir for your efforts Good luck with glades ❤❤❤

[@raya.pawley3563]

Thank you

[@JulesMoyaert_photo]

👍

[@ryanmartinez7213]

Yes! Algebra time!

[@jamestalbott4499]

Thank you!

[@Daniel-OConnell]

(b-d) + (c+d) = (b+c) = 11+7 =18. (b+c) - (a+b) = (c-a) = 18-9. Thus (c-a) =9 Thus factors a and c must be 12 & 3 both positive or both negative.(pos. Product) & (a+c) = 15 or -15 (a+c) + (c-a) =2c = 9+15=24 or 9 -15 =-6 . Thus c = 12 or - 3 and a = -12 or 3 Then if a =3, b=6, c=12, d=-5 or If a =-12, b=21 , c=-3 , d =10 ,

[@Tmwyl]

I got this one!

[@rcnayak_58]

We can still find the values in simpler ways. Now (a+b) - (c+d) = 9 -7 = 2. This can be rearranged as (a - c) + (b - d) = 2. We know b - d = 11. Putting this value, (a - c) = 2 - 11 = -9. We already know ac = 36. Using algebraic formula, (a + c)² = (a -c)² + 4ac, we get (a + c)² = (-9)² + 4 x 36 = 225, so that (a + c) = ± 15. Case - I, say (a + c) = 15 and we know (a - c) = -9. So our a = (15+(-9))/2 = 3 and c = (15 -(-9))/2 = 12. Since a + b = 9, b = 9 - a = (9 - 3) = 6. Similarly, d = 7 - c = 7 - 12 = -5. Therefore ( a, b, c, d = 3, 6, 12, -5). Case - II, (a + c) = -15 and (a - c) = -9. So, a = (-15+(-9))/2 = -12, c = (-15 - (-9))/2 = - 3, b = 9 - (-12) = 21 and d = 7 - c = 7 - (-3) = 10. The second set values are ( a, b, c, d = -12, 21, -3, 10).

[@joeschmo622]

✨Magic!✨

[@Viesto1980]

You can use only a and b. a*b=36 and (9-a)-(7-b)=11 => -a+b=9 => b=9+a. a(9+a)=36 =>a*2+9a-36=0. Solve this quadratic equation and you will get the same answer .

[@thisarasenadi2839]

❤❤❤❤❤

[@gabricia]

Dos soluciones: -12, 21, - 3 y 10 3, 6, 12 y - 5 Una -12 + 21 -3 + 10 Y la otra 3+6 12-5

[@CHIRANJITTTTTT]

use comma (,) ...! 🕉️🔱🇮🇳🚩🙏🚩🇮🇳🔱🕉️

[@jaimeyomayuza6140]

A=3 B=12 C=6 D=-5 A=-12 B=-3 C=21 D=10 desde Bogota D.C. Colombia

[@michaeldoerr5810]

I got the answer. The answer is (a,b,c,d)=(3,6,12,-5). I believe that this is the FIRST TIME that I got this puzzle right just following along your presentation without finishing the video. I am planning on watching similar videos on your channel and apply that strategy on a similar puzzle in the Math Hunter channel. I am planning on being a mathematician!!!

[@marcgriselhubert3915]

There is another solution...

[@marcokretschmer3550]

You forgot the second solution... (a,b,c,d) = (-12, 21, -3, 10)... But don't stop believin'... 😊

[@prossvay8744]

The first box=3 Second box=6 Third box=12 And fourth box=-5

[@DandoPorsaco-ho1zs]

Let the squares be a, b, c and d. b = 9 - a c = 36 / a d = b - 11 = 9 - a - 11 = -a - 2 c + d = 7 = (36 / a) + (- a - 2) = 7 Using the last equation: a² + 9a - 36 = 0 a = (-9 ± sqrt(81+4x36))/2 = (-9 ± 15)/2 a = -12, b = 21, c = -3, d = 10 or a = 3, b = 6, c = 12, d = -5

[@rabotaakk-nw9nm]

2:30 (4)+(3)-(1)=(c-a=9) c=a+9 -> (2) => a(a+9)=36 => => a²+9a-36=0 => a=(-12, 3) a=(-12, 3) -> (1) => b=(21, 6) a=(-12, 3) -> (2) => c=(-3, 12) c=(-3, 12) -> (4) => d=(10, -5)

[@wackojacko3962]

I overthunk this problem and filled the pretty pink boxes with numbers closer to the Quantum Scale becuz I didn't have any specific objects handy to complete the order. 🙂

[@alster724]

Here is an alternate method (a shorter way) a+b= 9 c+d= 7 ac= 12 b-d= 11 =========== a+b= 9 (c+d= 7) (-1) a+b-c-d= 2 Since b-d= 11,let's substitute this from the above equation a-c+11= 2 a-c= -9 a= c-9 Let's substitute this value in the below equation ac= 36 c(c-9)= 36 c²-9c= 36 c²-9c-36= 0 (c-12)(c+3) c= 12, -3 So there will be 2 cases, a positive and negative solution. CASE 1 (+) If c= 12, then a= 3, b= 6, d= -5 CASE 2 (-) If c= -3, then a= -12, b= 21, c= 10

[@MrPaulc222]

I started at a different place, got three wonderfully complicated looking fractional solutions, then the fourth was a mile out LOL.

[@calvinmasters6159]

Fun. It was clenched cheeks til the quadratic. When it wasn't a j-number... relief.

[@giuseppemalaguti435]

(x,y,z,t)=(3,6,12,-5)....=(-12,21,-3,10)...mah, spero che i calcoli siano corretti

[@mohammedminhajuddin]

and the another term also be same

[@mjimbooyah]

what about a= 4, b = 5, c = 9, and d = Minus 2

[@David-ee1pi]

b-d doesn't equal 11

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : System of 4 Equations and 4 Unknowns, knowing that : 01) A + B = 9 02) C + D = 7 03) A * C = 36 04) B - D = 11 05) A + B + C + D = 16 06) C + (B - 11) = 7 ; B + C = 18 07) A + (18 - C) = 9 09) A - C = - 9 10) C = A + 9 11) A * (A + 9) = 36 12) A^2 + 9A - 36 = 0 13) Two Solutions : A = - 12 or A = 3 14) B = 9 - A ; B = 21 or B = 6 15) C = 36 / A ; C = - 3 or C = 12 16) D = B - 11 ; D = 10 or D = - 5 ANSWERS : A = - 12 ; B = 21 ; C = - 3 ; D = 10 OR : A = 3 ; B = 6 ; C = 12 ; D = - 5

[@Birol731]

My way of solution ▶ a+b= 9 a*c= 36 c+d= 7 b-d= 11 ⇒ a+b= 9 b=9-a a*c= 36 c= 36/a c+d= 7 36/a + d= 7 36+ad= 7a 7a-ad= 36 a(7-d)= 36 a= 36/(7-d) ⇒ b-d= 11 b= 9-a a= 36/(7-d) ⇒ 9-[36/(7-d)] - d= 11 9+ 36/(d-7) - d = 11 both sides multiplied by (d-7) : 9(d-7) + 36 - d(d-7) = 11*(d-7) 9d-63 +36 -d²+7d= 11d-77 d²-5d-50=0 Δ= 25-4*1*(-50) Δ= 225 √Δ= 15 d₁= (5+15)/2 d₁= 10 ⇒ a₁= 36/(7-d) a₁= 36/(7-10) a₁= -12 c₁= 36/a₁ c₁= 36/(-12) c₁= -3 b₁= 9 - a₁ b₁= 9 - (-12) b₁= 21 II) d₂=(5-15)/2 d₂= -5 ⇒ a₂= 36/[7-(-5)] a₂= 36/12 a₂= 3 c₂= 36/a₂ c₂= 36/3 c₂= 12 b₂= 9 - a₂ b₂= 9 - 3 b₂= 6 So, we have 2 groups of solutions !

[@alster724]

This is how I did it... GIVEN a+b= 9 c+d= 7 ac= 36 b-d= 11 c+d= 7 b-d= 11 (b+c= 18) -1 a+b= 9 a-c= -9 ac= 36 a=36/c (36/c-c= -9)c 36-c²= -9c c²-9c-36 = 0 (c-12)(c+3) c= 12, -3 There are 2 solutions SOLUTION 1 If c= 12, then a= 3 b= 6 d= -5 SOLUTION 2 If c= -3 a= -12 b= 21 d= 10

[@mohammedminhajuddin]

bro it is so simple in the first line the sum of a and b is 9 and 9 is only written in two ways as the sumof two numbers 3+6 and 9+0 if we see the 9+0so the multiplication row must be zero but it should be 36 so we must take 3 and 6 as a and b if we know a and b we can easily find c by algebric expression and we got the all values intha

[@Daniel-OConnell]

and 1+8 , 2+7, 3+6, 4+5 etc . I don't quite understand your reasoning.

[@sergioaiex3966]

Solution: a + b = 9 ... ¹ a × c = 36 ... ² b - d = 11 ... ³ c + d = 7 ... ⁴ ¹ - ³ (a + b) - (b - d) = (9) - (11) a + b - b + d = - 2 a + d = - 2 ... ⁵ ⁵ - ⁴ (a + d) - (c + d) = (-2) - (7) a + d - c - d = - 9 a - c = - 9 ... ⁶ a × c = 36 ... ² a - c = - 9 ... ⁶ a = 36/c ... ⁷ 36/c - c = - 9 (× c) 36 - c² = - 9c c² - 9c - 36 = 0 c² - 12c + 3c - 36 = 0 c(c - 12) + 3(c - 12) = 0 (c - 12) . (c + 3) = 0 c = 12 c = -3 ========== a × c = 36 ... ² a × 12 = 36 a = 3 a × c = 36 ... ² a × (-3) = 36 ... ² a = -12 ========== ========== a + d = - 2 ... ⁵ 3 + d = - 2 d = - 5 a + d = - 2 ... ⁵ -12 + d = - 2 d = 10 ========== ========== b - d = 11 ... ³ b - (-5) = 11 ... ³ b + 5 = 11 b = 6 b - d = 11 ... ³ b - (10) = 11 b = 21 ========== So in this problem there are 2 solutions: a = 3 b = 6 c = 12 d = - 5 a = - 12 b = 21 c = - 3 d = 10

[@unknownidentity2846]

No geometry? No problem: . .. ... .... ..... a + b = 9 c + d = 7 a * c = 36 b − d = 11 ⇒ d = b − 11 a + b = 9 c + (b − 11) = 7 ⇒ b = 18 − c a * c = 36 a + (18 − c) = 9 ⇒ c = a + 9 a * c = 36 a * (a + 9) = 36 a² + 9*a − 36 = 0 ⇒ a = −9/2 ± √[(9/2)² + 36] = −9/2 ± √(81/4 + 144/4) = −9/2 ± √(225/4) = −9/2 ± 15/2 First solution: a = −9/2 + 15/2 = 6/2 = 3 b = 18 − c = 18 − 12 = 6 c = a + 9 = 3 + 9 = 12 d = b − 11 = 6 − 11 = −5 a + b = 3 + 6 = 9 ✓ c + d = 12 + (−5) = 7 ✓ a * c = 3 * 12 = 36 ✓ b − d = 6 − (−5) = 11 ✓ Second solution: a = −9/2 − 15/2 = −24/2 = −12 b = 18 − c = 18 − (−3) = 21 c = a + 9 = −12 + 9 = −3 d = b − 11 = 21 − 11 = 10 a + b = −12 + 21 = 9 ✓ c + d = −3 + 10 = 7 ✓ a * c = (−12) * (−3) = 36 ✓ b − d = 21 − 10 = 11 ✓ So we have two valid solutions: 3 6 12 −5 −12 21 −3 10 Best regards from Germany

[@davidshen5916]

1️⃣-4️⃣-3️⃣ A-C=-9， A*C=36，A=3，C=12，A=-12，C=-3