You see nonlinear equations, they see linear algebra! (Harvard-MIT math tournament)

[@blackpenredpen]

Get started with a 30-day free trial on Brilliant: 👉brilliant.org/blackpenredpen/ ( 20% off with this link!)

[@ibrahimhalilkanat299]

you are talking so much

[@nzbil8790]

i have another way xy + z + xz + y = 40 + 51 x(y+z) + y + z = 91 (x+1)(y+z) = 91, we know that y+z = 19 - x, so : (x+1)(19-x) = 91 19x - x^2 + 19 - x = 91 x^2 - 18x + 72 = 0 (x-12)(x-6) = 0 x = 12 or 6 then just plug x into the equation and we will get the same value of y and z like in the video, CMIIW :3

[@NarutoSSj6]

Yea. He said he would leave the substitute method for us to use. This was about using linear algebra to solve the problem

[@nzbil8790]

@@NarutoSSj6 okay

[@dummyaccount1706]

​@@NarutoSSj6no way, did he really?

[@ПерстФомы]

Я решил так же😄. I've decided the same way

[@Nothingx303]

Same bro 😊

[@harshitgautam4436]

Add 1st and second equation (y+z)(x+1)=91 x+1+y+z=20 Take x+1 as a and y+z as b Then ab=91 and a+b=20 So (a,b)=(13,7),(7,13) x is either 6 or 12

[@vuqarahadli]

yep i did it in my mind in 1 minute

[@DistortedV12]

As they say, “linear algebra is limited to linearity property, but oftentimes, it is easy to fit your nonlinear problem in a linear form”

[@mattywlion5174]

Who says that? The voices in the head?😂

[@Steve_Stowers]

I started Gauss-Jordan elimination on the matrix at 5:18, got to [1 1 19-x] [0 x-1 x+32] [0 1-x x²-19x+40] and concluded that -(x+32) = x²-19+40 (to make row 2 + row 3 = 0), which gives x=6 and x=12.

[@landsgevaer]

Adding first two gives (x+1)(y+z) = 91 Final gives (x+1)+(y+z) = 20 Solving for x+1 and y+z gives the pair (7, 13), in either order. Backsubstituting in the first eqns, et voila.

[@joseluishablutzelaceijas928]

Yeah... also I did it in a very similar way but I then interpreted "x+1" and "y+z" as the zeroes of the equation z^2-20*z+91 = 0 and obtained so the two solutions suggested in the video, this problem can thus be solved quite quickly, maybe he simply wanted to intentionally apply concepts from linear algebra here, although that would not be necessary.

[@Bayerwaldler]

Only under the assumption that x+1 is an integer

[@landsgevaer]

@@Bayerwaldler Nope, why? Solving for x+1 and y+z allows them to be any real numbers (or complex or much more). That is a quadratic. It just turns out to be an an integer solution.

[@landsgevaer]

@@joseluishablutzelaceijas928 Yes, I agree. I give a LinAlg course this quarter and thought of borrowing it as a different type of exercise when I saw the title, but I think it is a bit contrived given that easier approaches come to mind.

[@Bayerwaldler]

@@landsgevaer Aah - I see. You get a quadratic equation like -u^2 + 20u = 91 where u = x+1. Then the rest follows nicely. Very good!

[@eqyuio654]

I have another way x + y + z = 19 xz + y = 51 xy + z = 40 y = 51 - xz z = 40 - xy x + 51 - xz + 40 - xy = 19 x - xz - xy = -72 -x (y + z - 1) = -72 x (y + z - 1) = 72 x + y + z = 19 then y + z - 1 = 18 - x x (18 - x) = 72 18x - x² - 72 = 0 x² - 18 x + 72 = 0 (x - 6) (x - 12) = 0 x1 = 6 x2 = 12 Next: y + z = 13 y + 6z = 51 6y + z = 40 6y + z = 40 3y + 3z = 39 3y + 18z = 153 (Substracting) -20z = -152 z1 = 7,6 y1 = 5,4 y + z = 7 y + 12z = 51 12y + z = 40 12y + z = 40 6y + 6z = 42 6y + 72 z = 306 (Substracting) -77z = -308 z2 = 4 y2 = 3 Solutions: x1 = 6 y1 = 5,4 z1 = 7,6 x2 = 12 y2 = 3 z2 = 4

[@lostwizard]

I combined the first two equations, did some algebra to get (x+1)(y+z)=91, used the third to replace (y+z) with (19-x), and arrived at that same quadratic. After which I promptly made a sequence of arithmetic errors solving the quadratic and then subsequently even more arithmetic errors solving the resulting system of equations. It turns out that if you fail at basic arithmetic, math is hard. :)

[@khattab5351]

it is actually solvable using gaussian elimination 0 x 1 | 40 0 1 x | 51 1 1 1 | 19 replace r1 and r3 r3-xr2 r1-r2 r2-xr3 r1-(1-x)r3 doing these steps gives and identity matrix and gives us x, y and z with respect to x x=(51x-40)/x+1 -32 then we get a quadratic equation that gives x=6 or 12 y=51- 51x^2-40x/x^2-1 z=(40-51x)/(1-x^2) we can then substitute 6 and 12 into these and get y and z s1 (6, 5.4, 7.6) s2(12, 3 ,4)

[@kontiki7030]

2:32 was is a coincidence because 5y+2z=10,y+z=3 and 2y-z=1 has a solution which is (y,z)=(4/3,5/3)

[@zyc4nthropy728]

2:43 I love how these 3 equations actually do still have a valid real number solution

[@matheusjahnke8643]

Time for cheap tricks; Computing L1+L2-L3- x L3 on the left hand side... we have (xy+z)+(xz+y)-(x+y+z)-x(x+y+z) carefully cancelling...we are left with -x-x² The right hand side is 40+51-19-19x=72-19x -x-x²=72-19x Then I got lazy... but we arrive at the same polynomial of x(up to a non-zero scalar multiple) so it should arrive at the same solution

[@milanstevic8424]

Here's another way: ...... then I got lazy... but we arrive to the end of the video anyway, so x=6, y=5.4, z=7.6 and x=12, y=3, z=4

[@felipefred1279]

I was doing this method of eigenvalues to solve systems of differential equations at my differential equations class today. Good lecture professor!

[@victorpaesplinio2865]

In the example at 2:30, the 3rd equation is actually the 1st minus 3 times the 2nd, meaning it is also redundant

[@joshuanugentfitnessjourney3342]

Linear algebra is one thevmost useful math tools i never used unless forced to in my undergrad

[@darcash1738]

What applications do you use it for now

[@joshuanugentfitnessjourney3342]

@@darcash1738 nothing, couldn't get a job with my math degree. Really useful in physics and especially computer science

[@sanjogar]

Adding the first two equations, you get (y + z)(1 + x) = 91. From the second equation: y + z = 19 - x. Substitute and get (19 - x)(1 + x) = 91. Solving for x with Bhaskara's Formula, you obtain x = 12 or x = 6, and so on.

[@jimschneider799]

How much would you bet that whoever came up with this problem started with the functions x*y+z, x*z+y, and x+y+z, picked some values for x, y, and z, and plugged them in to get the values x*y + z = 40, x*z + y = 51, and x+y+z = 19, only to be surprised later when it had multiple solutions?

[@janami-dharmam]

quadratic equations are expected to have two solutions; because the numbers are real, both roots must be either real or complex conjugate

[@iabervon]

Your digression about the 0 determinant not always giving solutions also applies in reverse. If you look at where the first two equations are linearly dependant, that must give 0 for the determinant, which gives you a root of the cubic (x=1) without needing to notice anything about the coefficients.

[@devondevon4366]

3, 4, and 12 This is a simpleproblem for an Harvard MIT math tournament xy+ z =40 equation 1 xz + y =51 equation 2 x+ y+z = 19 equation 3 Let's add the first two equations : xy + z =40 and xz + y =51, Hence xy + xz + y + z =91 x(y+z) + 1(y+z) =91 factor out y +z Equation 5 y + z = 19- x (solving for y+ z , using equation 3) x(19-x) + 1 (19-x) = 91 (substituting 19-x into equation 5) (x+1)(19-x) =91 (x+1)(-x+19)=91 - x^2 + 19- x + 19x =91 0 = x^2 -18x +72 0= (x -6)(x-12) x =6 and x =12 Let try x=12 first using equation 1 and equation 2 12y + z=40 y + 12z=51 y = 3, and z=4 Hence x=12 , y=3 and z= 4 The next step is to solve when x =6 and using equation 1 and equation 2 again 6y + z =40 equation 9 y + 6z= 51 equation 10 36y + 6z = 240 mulltiply equation 9 by 6 y + 6z = 51 35y = 189 y= 189/35 y=5.4 z=266/35 z=7.6 x =6, y=5.4 , and z=7.6 answer as well This also satisfy the equation when x= 6, Hence x=12 Hence the answer is x=12, y=3 and z=4

[@TalkLoudSayNothing]

You don't need to solve the cubic, you can just say that it's easy to see that the first two rows of the 3x3 system of equations are linearly independent, therefore the third row must be expressible as a linear combination of the first two. Then looking at the left hand side, the respective coefficients must be 1/(1+x) and 1/(1+x), which yields 91/(1+x) = 19-x, x=6 or x=12.

[@Qermaq]

2:42 y = 4/3, z = 5/3. The third equation works just fine....

[@cdkw8254]

I haven't heard about any of those things but it sounds cool.

[@prodbyKamikaZ]

I love how the math you do is simple and understandable, but used creatively.

[@kiopa5233]

Hey I am an high schooler, I didn’t know most of the determinant formulas you used , but your video was fascinating. Thank you.

[@mjkhoi6961]

quicker way to get x=6 and x=12: 1. add the two top rows together and factor to get (x+1)(y+z)=91 2. rearrange third row and substitute, y+z=19-x → (x+1)(19-x)=91 3. expand and factor to get (x-6)(x-12)=0 avoids having to compute a determinant and factor a cubic, and the inconsistent x=1 solution that the cubic provides

[@TheBlueboyRuhan]

THANK YOU for this idea, can't believe is was that simple to fix x as a constant and take determinants - since x is non-zero! Ignore all of the comments talking about the trivial way to solve this with gauss-elim, they don't understand the power this method has

[@jmsaucedo]

Just watch Gilbert strang course too

[@lawrencejelsma8118]

It was great seeing an extraneous solution that didn't solve the systems of those nonlinear equations by trying to turn a 3x2 matrix multiplied by a 2x1 matrix to form a 3x1 matrix of solutions. We saw extraneous x=1 that didn't work with x=6 and x=12 that both do work as solutions in what was taught. Sometimes looking at a problem creating polynomial solutions both extraneous and good solutions get intermixed to allow us to see potentials that fail but can inter result. 👍

[@matheodaniloalvitreslopez3159]

This question is to be answered until 11:59 am, for those who have the last digit of their DNI 5: What is the letter you like the most among A, B, C, I, N , The m?

[@72kyle]

First eq take last eq: xy - x - y = 21 (IV) Rearrange third z = 19-x-y Sub this into second x(19-x-y) + y = 51 (V) Expand this and add to (IV) 19x - x^2 -xy +y +xy -x -y = 21+51 Simplify to x^2 -18x +72 =0 (x-6)(x-12)=0 x=6, or x=12 Then sub into (IV) to get y and then into third eq to get z. (6,5.4,7.6) and (12,3,4)

[@janda1258]

Set up the equation x(Eq3) + (Eq3), and simplify using (Eq1) and (Eq2), you get x^2 -18x +72 = 0, then solve for x. Next, multiply (Eq1) by x, simplify using (Eq2), solve for y to get y=(40x-51)/(x^2-1) and solve for y. Lastly use (Eq3) to solve for z

[@wafiklotfallah9951]

Subtracting Eq 3 from Eq 1 and adding 1: xy-x-y+1=(x-1)(y-1)=22 Subtracting Eq 2 from Eq 1 and adding 1: xz-x-z+1=(x-1)(z-1)=33 Substituting X=x-1,etc., we get: XYZ=22Z=33Y Or Z=3Y/2 X+Y+Z=16. So 2X+5Y=32 Subtracting Eq 1 from Eq 2: zX-yX=(Z-Y)X=11. So XY=22 Now we get: (2X-5Y)^2 = (2X+5Y)^2 - 40XY = 1024 - 880 = 144 So 2X-5Y=+/- 12 4X = 44,20. We then get X, Y, Z, and finally get the two solutions: x=12,y=3,z=4 or x=6,y=27/5,z=38/5

[@twelfthdoc]

Before looking at BPRP's solution: I was able to factor the information in the equations into a quadratic in terms of x, yielding two results. One case yielded a solution in the Rationals while the other yielded a solution in the Naturals, so depending on the set the variables are meant to be from, there are either one or two solutions. After watching BPRP's solution: the quadratic I obtained was the same, but the factor of (x - 1) in the cubic was a redundant solution because of the matrix multiplication. I eliminated that solution because I had used a linear transformation while I was manipulating the original equations to extract the quadratic in x. Always worth checking solutions with the original equations as well as any obtained through linear algebra manipulation to ensure they are true solutions rather than redundant (i.e contradictory) solutions.

[@BalajiKomanabelli-nd1xq]

Loved this approach, thinking out of the box

[@Chris_387]

Add the first 2 equations, factor and take 2 cases, you are done

[@danieldepaula6930]

If I understood correctly, that's the same way I did it. If you add the first two equations, you get x(y+z) + (y+z) = 91. After that, using the 3rd equation, you get y+z = 19 - x. So, you can easily find both values ​​of x. Then, using the first two equations again (this time, as a linear system of order 2), you find y and z for both cases.

[@Bayerwaldler]

Only under the assumption that x+1 is an integer

[@ZackBlackwood97]

​@danieldepaula6930 genuinely asking as I'm not 100% sure, when you add the first two equations together don't you need to do X×Z?

[@danieldepaula6930]

​​@@ZackBlackwood97 I didn't understand exactly what you mean, but this is it (step by step): xy + z + xz + y = 40 + 51 xy + xz + y + z = 91 x*(y+z) + (y+z) = 91 And as you have that y+z = 19-x by the third equation, you can find x.

[@djconnel]

@@danieldepaula6930but that has complex roots

[@zeron85]

Proud to have been able to solve this problem

[@TheJaguar1983]

When I was in school these were called "Simultaneous Equations". Not sure when the term "Systems of Equations" became the preferred term. I only heard for the first time about a year ago.

[@oloyt6844]

Extremely interchangeable but system usually refers to 3+ variables in my experience

[@pneujai]

‘simultaneous equations’ is commonly used in school mathematics ‘system of equations’ is commonly used in the field of linear algebra, from high school to university

[@christianbarnay2499]

7:43 You can immediately see there's a X-1 factor in the determinant and factor it right away before expanding the 19-X. X²(19-X)+51+40-(19-X)-51X-40X = (X²-1)(19-X)+51(1-X)+40(1-X) = (X-1)(X+1)(19-X)-91(X-1) = (X-1)((X+1)(19-X)-91) = (X-1)(-X²+18X-72)

[@rcnayak_58]

We can also solve it without using determinant. For example, let us arrange these equations as xz + y = 51 ...(1), xy + z = 40 ...(2) and x + y + z = 19 ...(3). Adding (1) and (2) and factorizing, we get (x + 1)(z + y ) = 91 ... (4). Again subtracting (2) from (1), we get , on factorizing, (z - y)(x - 1) = 11 ...(5). From (3), (z + y) = 19 - x ...(6). Putting (z + x ) value in (4) , we get (x + 1)(19 - x) = 91 ... (7) solving this , we get ( x - 6)(x -12) =0, so that we have x = 6 or x = 12. Case - I, when x = 6, in (3), we we get z + y = 13 ... (7) and again putting x =6 in (5), we get (z - y) = 11/5 ... (8). From (7) and (8) we get z = (1/2)(13+11/5) = 7.6 and y = (1/2)(13 - 11/5) = 5.4. Therefore, the first solution set {x, y, z) = {6, 5.4, 7.6}. Case - II when x =12: putting this in (3) z + y = 7 ...(9). Again putting x = 12 in (5) we get z -y = 1 ...(10). From (9) and (10), we get z = (1/2)( 7 + 1) = 4 and y = (1/2)( 7 - 1) = 3. Therefore, the 2nd solution set {x, y, z) = {12, 3, 4}.

[@marusukech.5049]

Another way is let w = y + z and it can form an quadratic equation

[@CalmArgha1243]

The way I had solved this I got values which when added and multiplied givens answers close to 40,51,19. The set of answers was (x,y,z)=(7.3,4.6,6.2) and this is the only set I got.

[@hai.nguyen995]

I guess I over-complicated this: xz + y - (x - y - z) = 32 => xz - x - z +1 = 33 => (x-1)(z-1) = 33 (1) xz + y - xy - z = 11 => (x-1)(z-y) = 11 (2) 1/2 => z - 1 = 3(z - y) => z = (3y - 1)/2 => x = (39 - 5y)/2 xy + z = 40 => 5y^2 - 42y + 81 = 0; y = 3 or y = 27/5 (x,y,z) = (6, 27/5, 38/5) or (12, 3, 4)

[@Protract_Loop]

Just subtract eq.3 from eq.2 = eq.4 Subtract eq.4 from eq.1=eq.5 Substitute the value of x from eq.3 in eq.5 Factorise eq.5 and you get value of x Substitute value of x and y in eq.1 to get y and z

[@abhinavyadav8408]

we can solve by adding the equations xy+z=40 eq1 xz+y=51 eq2 x+y+z=19 eq3 on adding eq. 1,2,3 x+xy+xz+2y+2z=110 adding 2 on both the sides x(y+z+1) 2(y+z+1)=112 (x+2) (y+z+1)=112 -------equation 4 from equation 3 y+z=19-x substituting y+z=19-x in eq. 4. (x+2) (-x+20)=112 -x^2+18x-72=0 (x-6)(-x+12)=0--------equation 5 from equation 5 x=6, x=12 substituting x=6 in equations 1,2,3 we get 6y+z=40--eq6 6z+y=51---eq7 6+y+z=19 --eq8 from equation 8 y+z=13 from equation 6 z=40-6y substituting z=40-6y in eq. 8 -5y=-27 y=5.4 substituting y=5.4 in equation 8 z=7.6

[@lucho2868]

Suming the first two equations gives (x+1)(y+z) = 91 and the third equation gives x+1+y+z=20 thus x+1 = 10 +/- sqrt(100-91) = {13,7}. In the first case (x=12) we get (Cramer) y=(40x-51)/143=429/143=3 and z=(51-3)/12=4. In the second case (x=6) we get (Cramer) y=(6×40-51)/35=189/35=27/5 so z=40-27*6/5=38/5. So, after checking they work, the set of solutions is {(12,3,4),(6,27/5,38/5)}. EZ but tedious.

[@davidturner9827]

(40 + 51) / (x + 1) = 19 - x

[@Anonymous-tq2iu]

6:16 determinant of a determinant ?

[@MarieAnne.]

Solve: (1) xy + z = 40 (2) y + xz = 51 (3) x + y + z = 19 First we solve for y and z in terms of x (1)−(3) xy − x − y = 21 → y = (x+21)/(x−1) (2)−(3) xz − x − z = 32 → z = (x+32)/(x−1) Next we plug into (3) to solve for x x + (x+21)/(x−1) + (x+32)/(x−1) = 19 x(x−1) + (x+21) + (x+32) = 19(x−1) x² − x + 2x + 53 = 19x − 19 x² − 18x + 72 = 0 (x−6) (x−12) = 0 x = 6, x = 12 Then we use value of x to find y and z. x = 6, y = (6+21)/(6−1) = 27/5 = 5.4, z = (6+32)/(6−1) = 38/5 = 7.6 x = 6, y = (12+21)/(12−1) = 33/11 = 3, z = (12+32)/(2−1) = 44/11 = 4 Solutions: *(x, y, z) = (6, 5.4, 7.6) or (12, 3, 4)*

[@nirorit]

I multiplied the 3rd by x, and then substituted xy and xz with the first 2 equations and then substituted y+z with the 3rd equation to get a quadratic equation with only x.

[@ΕκπαιδευτήριαΚαντά-η1ω]

det(x 1, 1 x)=x^2-1 detψ(40 1,51 x ) =40x-51 detz(χ 40,1 51)=51χ-40 ψ=detψ /det z=detz/det we substitute into third equation x^3-19x^2 +90x-72=0 x1 x=6 ψ=5.4 z=7.6 x=12 ψ=3 z=4 x>=1 x

[@carlosangulo2888]

Love the way u solve it. Thx. ❤

[@curtiswfranks]

For those who are saying to do it a different way: Often, your method is limited to basically this exact problem. If we change this problem so that the coefficients or constants are different, your technique becomes cumbersome. But this technique will continue to work without modification. The solutions may be gross, but it nonetheless will still obtain them.

[@nyaanyaa7]

xy+(19-x-y)=40 x(19-x-y)+y=51 adding both eq to get 18x-x^2+19=91 so x^2-18x+72=0 then x=6, x=12 solve for y and z and get 2 solutions 12,3,4 and 6,27/5,38/5

[@rogo7330]

Basically, we assume that this system is solvable, and from that assumption we solving simpler system, avoiding undefined behaviour, like setting variable to zero.

[@mikezilberbrand1663]

(x+1)*(y+z)=91 - add the first two. x+1 + y+z = 20 Solve the quadratic.

[@paultoutounji3582]

You could have just added equations 1 and 2 and factorise as (x+1)(y+z) = 91. Then substitute from equation 3 y+z = 19-x…..

[@Maman-Setrum]

wow, i like how you using matrix as part of solving

[@manavverma55]

Dude I have similar question for you to solve x+y+z=4, xy+yz+zx= -7, xyz= -10 find x, y & z. tell me in another way or easy way for this question i try substitution but it was became very long and lengthy.

[@matheodaniloalvitreslopez3159]

Esta pregunta es para que me respondan hasta las 11:59 am, para los que tienen el último dígito de su DNI 5 : ¿Cuál es la letra que les gusta más entre la A, la B, la C, la I, la N, la M?

[@hamzaansari1772]

Surprisingly Seeing the third eqn first time i randomly put values (x,y,z) as (12,3,4) and ❤ question is over But answer is not the thing that mathematics wants😊 Now trying with the original method

[@__NK_37]

just add equation first and second and factorise

[@cringotopia8850]

Truely useful method that gave me new insights about using the traditional Gaussian Elimination However, If I were to be in that tournament, i would use the traditional Substitution method, because it seems specifically easier and faster in this question

[@hpsmash77]

11:55 that transition

[@michaelmann8800]

Far easier to combine the first two equations, factor it, and then use the third equation to write the combination of the first two in terms of x alone.

[@tharunsankar4926]

I assumed that x had to control the rank of the system. I determined that the rank had to be 2, so I did Converted the equation to augmented matrix system (4 x 3) and determined the x value to ensure that one of the rows go to 0’s for rank 2. I got I think 2 values for x (I have to check I did this a long time ago) that trip this equation to a rank 2. Then solved for y a and x using cramers rule. Anyways moral of the story: I don’t think you have to get the characteristic polynomial to trip this into a diminished rank anyways.

[@ChessBros-ic5tz]

What happens if you equate phytagoras Therom to the quadratic formula or is that not possible

[@ethanthiebaut2596]

More of these please 🙏🙏

[@spiderjerusalem4009]

I thought you used the matrix approach only on the first 2 eq ⎡x 1⎤⎡y⎤ ⎡40⎤ ⎣1 x⎦⎣z⎦ = ⎣51⎦ obtaining (y,z)=(40x-51,51x-40)/(x²-1) and then just substitute on third eq

[@mauriziomorales5303]

Beautyfull. Thank you very much because you helped me to solver this exercise of Lineal Algebra.

[@joellouis1005]

I just think of an easy way which high school can solve this. Since x +y +z = 19, z = 19 -x -y ; xy +z = 40, so xy + 19 -x -y = 40, then xy -x -y =21 … (1) xz + y = 51, so x (19 -x -y) +y = 51, then 19x -x^2 -xy +y = 51 …(2) Put (2) into (1), -x^2 +19x -xy +x +y -x =51, so -x^2 +18x -21 =51, x^2 -18x +72 = 0. We can get x = 6 or 12

[@januszkorwin-mikke7277]

I am oldschool. I do not count. I think. Suppose x, y, z are N. Then. From [3] either one is odd or all three are odd. From [2] all cannot be odd and the one odd is y. If y=1 then [1] & [3] contradict. If y=3 then from [1]&[3] x=12 and z=4. I am lucky - am I not? It is called intuition... If y=5 then x=13/2 not-N, contradict [2]. If y=7 then x=14/3 not-N contradict [2]. If y=9 then x=15/4 not-N contradict [2]. If y=11 then x=16/5 not-N contradict [2]. If y=13 then x=1/2 not-N contradict [2]. If y=15 then x=2/7 not-N contradict [2]. If y=17 then x=1/8 not-N contradict [2]. If y=19 then x=20/9 contradict [2]. All done in head in 2 mins. Probably there is also one non-N solution, but I am going to bed. O! @nzbil8790 is better, using no intuition

[@ahmetalicetin5331]

Can you make a video on how to use Weierstrass factorization theorem?

[@graikoyt]

(xy + z) + (xz + y) - (x + y + z) = 40 + 51 + 19 xy + xz - x = 72 x(y + z - 1) = 72 x(19 - x - 1) = 72 18x - x^2 = 72 x^2 - 18x + 72 = 0 x = 6 or 12 When x=6 or 12, you can solve for y and z using first two equations Solutions are (6,27/5,38/5) and (12,3, 4)

[@alexmucci5327]

So, we're basically treating the X as a parameter and using the standard method for linear system? Well, pretty genious move; good job americans

[@armanavagyan1876]

Pretty interesting PROF 👍

[@nuclearrambo3167]

i was thinking of row reduction while treating x as const

[@scottleung9587]

Got 'em both!

[@MichaelZankel]

Yess I asked for more last video and got more 😂

[@notfancy2000]

I haven't seen Ruffini in ages!

[@prateek1.9]

add all eqns , we get xy + z + xz + y + x + y + z = 110 xy + xz + x + 2y + 2z = 112 - 2 x(y + z + 1) + 2y + 2z + 2 = 112 x(y + z + 1) + 2(y + z + 1) = 112 (y + z + 1)(x + 2) = 112 (y + z + 1)(x + 2) = 8 * 14 x + 2 = 14, y + z + 1 = 8 x = 12..........(a), y + z = 7.........(b) given xy + z = 40 from eq(a), 12y + z = 40 11y + (y + z) = 40 from eq(b), 11y = 40 - 7 11y = 33 y= 3.........(c) also given, x + y + z = 19 from eq(a) and (c), 12 + 3 + z = 19 z = 4 (x,y,z) = (12,3,4) integer soln

[@kacodemonio]

This is a beautiful problem.

[@ianmathwiz7]

If we take our solutions to be in projective space, the x=1 case provides one more solution at infinity: (1, 0, 0, 0).

[@yoav613]

Nice and very easy

[@mehdimosbah3654]

xy+z-x-y-z =21 x(y-1)-y=21 x(y-1)-y+1=22 (x-1)(y-1)=22 x=12, y=3 z=19-15=4 (12,3,4)

[@RisetotheEquation]

The most important thing to remember: MIT >> Harvard

[@spenzr6920]

Add first two equations and put y+z = 19-x

[@ulysslombu-dji-mabicke1868]

Nice. I didn't know that trick.

[@danielmilyutin9914]

What questiined me is why did value 1 appeared? After a thought, beacuse it makes matrix singular not with row 3 being linear combination of rows 1 and 2, but columns 1,2 became collinear.

[@cillixnlynch_]

Could also use multivariable Newton method

[@Why-cl8pf]

Solved this in the real test! :)

[@Nine-2545]

I have a question, Let A B C, and k are Real numbers. If 2k^2k = ((A+B+C)^(A+B+C))/2, find k in terms of A, B, and C.

[@fernandolino6493]

12,4,3

[@ramunasstulga8264]

Mit never disappoints 🗿

[@kdenis8852]

Did it in my head, 12,4,3

[@jamescollier3]

I'd stop at the matrix.... then the cubic 😅 [edit] then for sure the 7.6.... lol

[@ahmedshaikha8938]

Ever heard of the Moore Penrose inverse

[@HenryBriskin]

This man is a genius

[@dimokratisnt3637]

Any closed contour integral videos?

[@chunpanhuen2681]

It is a really good question

[@animeaura2703]

I did like this and got this answer: XY+Z + 11=XZ+Y XY-XZ + Z-Y=-11 X(Y-Z) -1(Y-Z)=-11 (Y-Z) (X-1)=-11 so, Y= -11+Z and. X= -10 THEN, X+Y+Z=19 -10 -11+Z+Z=19 2Z=40 Z=20 Y=-11+20 = 9 Therefore, X= -10, Y= 9 and Z= 20

[@kenchen1351]

How do you go from "(Y-Z) (X-1)=-11" to "Y= -11+Z and. X= -10" ?

[@animeaura2703]

@@kenchen1351 Wdym by how did I go from here to there? It's just (Y-Z) (X-1) = -11 So Either, or, Y-Z = -11 X-1 = 11 or, Y= -11-Z therefore, X = -11+1= -10. Any questions??

[@kenchen1351]

@@animeaura2703 There is no rule specify that X,Y or Z is integer. In addition, even if it is integer. Y-Z could be 1, -1, 11, or -11. X-1 could also be one of the 4 numbers.

[@animeaura2703]

@@kenchen1351 Then is my answer also one of the solutions or not ? ( I'm just a 9th standard kid so I don't know much)

[@kenchen1351]

@@animeaura2703 You can double check your answer by putting your answer back to the question: For example, XY + Z = 40. When put in X= -10, Y= 9 and Z= 20. You get (-10)*9 + 20 which is -70 instead of 40. Therefore, the answer is incorrect.