It's a small application of Cramer's rule. The rule is that in the system Ax=b, the value of x_i (i=1,...,n) in *x* is equivalent to the determinant of A_i over the determinant of A, where A_i is the matrix A with the ith column substituted with the *b* vector. The simplest way to understand the proof is that a simultaneous system of equations always gives an invertible coefficient matrix A. Take X_i = I (the identity matrix) with *x* substituted into the ith column -- this is equivalent to saying, for example, that X_1 has the columns A⁻¹b, A⁻¹v_2, A⁻¹v_2, ... , A⁻¹v_n, where v_n is the nth column of A. Remembering that A_i is A with *b* in the ith column, we can describe A_1 having columns b, v_2, ... , v_n, and therefore X_1 = A⁻¹A_1, and because determinant is a linear operation and det(A⁻¹)=1/det(A), assuming A is an invertible matrix, we get x_1 = det(X_1) = det(A⁻¹A_1) = det(A⁻¹)det(A_1)= det(A_1)/det(A) QED That's the proof we did in linear numerical analysis, at least, there may be easier ones out there. Anywho, when you apply it to this small case of A being 2x2 (a system of two variables), you get the matrices A = [3 2] , x = [x] , b = [18] [2 5] [y] [23] and using Cramer's rule, we get | 18 2 | | 23 5 | 18*5 - 23*2 x = _________ = _______________ = 44/11 = 4 | 3 2 | 3*5 - 2*2 | 2 5 | and you could just plug x back in like they do in the video, or you could keep using Cramer's rule for x_2 (a.k.a. y), this time subbing *b* into the second column of A: | 3 18 | | 2 23 | 3*23 - 2*18 y = _________ = _______________ = 33/11 = 3 | 3 2 | 3*5 - 2*2 | 2 5 | giving us the correct x=4, y=3!

Cramer's rule is fine, but overkill for this small system. You can see it as just the elimination method with the algebra steps already done, kind of like the quadratic formula is just completing the square with the algebra done for you

@@SimonTyler0Could you explain what the “elimination method“ is? I don’t know it and so the rest didn’t follow for me. Thanks.

​@@jballenger9240 This is the elimination method: 3x+2y=18....(equation 1) 2x+5y=23....(equation 2) Suppose we want to eliminate y and find the value of x first, then multiply equation 1 by the coefficient of y from equation 2 and vice versa. So we get the two new equations as: 15x + 10y = 90...(equation 3) 4x +10y = 46...(equation 4) Now you can see that the coefficients of y in both the equations are equal, so now we just need to eliminate y term by subtracting one of the equation with the other, let's subtract equation 3 from equation 4. 4x +10y = 46 - (15x + 10y = 90) i.e. 4x + 10y = 46 -15x -10y = -90 And we get: -11x + 0 = -44 Therefore, 11x = 44 Therefore, x = 4 Now simply substitute the value of x in any of the original two equations to get y. Let us substitute the value of x in equation 1. So, 3(4) + 2y = 18 2y = 18-12 2y = 6 y = 3 In the video, to find x they multiplied 18 by 5 and 23 by 2 and subtracted it, which is also what is done in elimination method. They then divided it by 15-4 which is basically coefficients of x multiplied with the other equations' y coefficients. Try the elimination method and figure out how it works, then you can understand how the method used in the video is just a short form of the elimination method.

Thought the same as you. I teach Algebra 2 and this is just Cramer's Rule presented in a different way.

I'd definately teach how to do these equations the usual way first.

@@tecmath Why? What makes doing something like system-of-linear-equations or Cramer's rule "more" correct or "more" fundamental than this? 🤔

@@I.____.....__...__I think It's because the way how one particular maths topic is taught should be one that is likely to lead towards intuitive understanding. Applying clever tricks and shortcuts are certainly a distinct part of the big picture of Mathematics, but very often just quite difficult to see it through and to really know what's going on and why. Sometimes though, the tricky shortcut is what is actually taught in schools. Like dividing fractions for example.I guess it's because it's a hell-lot easier to teach it to kids compared to the proper and long winded way of doing the division of fractions. And it also takes up a lot less space on paper.

Yes, it is.

No doubt! If my k-12 teachers were as good as he is I wouldn’t have struggled nearly as much as I did

That's because it is.

Correction *The matrice [a x c y]

Thank you Sarah. Hope you liked it.

Do it 2 by 2

Same principle, it's still an Ax=b system of equations using Cramer's rule, just with three variables and a 3x3 A matrix. [ 2 1 1 ] [ x ] [ 8 ] A = [ 5 -3 2 ] , x = [ y ] , b = [ 3 ] [ 7 1 3 ] [ z ] [ 20 ] so | 8 1 1 | | 3 -3 2 | det(A1) | 20 1 3 | 8*(-3*3 - 1*2) - 1*(3*3 - 20*2) + 1*(3*1 - 20*-3) x = ________ = _____________ = _____________________________________________ = 6/3 = 2 det(A) | 2 1 1 | 2*(-3*3 - 1*2) - 1*(5*3 - 7*2) + 1*(5*1 - 7*-3) | 5 -3 2 | | 7 1 3 | | 2 8 1 | | 5 3 2 | det(A2) | 7 20 3 | 2*(3*3 - 20*2) - 8*(5*3 - 7*2) + 1*(5*20 - 7*3) y = ________ = _____________ = _____________________________________________ = 9/3 = 3 det(A) | 2 1 1 | 2*(-3*3 - 1*2) - 1*(5*3 - 7*2) + 1*(5*1 - 7*-3) | 5 -3 2 | | 7 1 3 | | 2 1 8 | | 5 -3 3 | det(A3) | 7 1 20 | 2*(-3*20 - 1*3) - 1*(5*20 - 7*3) + 8*(5*1 - 7*-3) z = ________ = _____________ = _____________________________________________ = 3/3 = 1 det(A) | 2 1 1 | 2*(-3*3 - 1*2) - 1*(5*3 - 7*2) + 1*(5*1 - 7*-3) | 5 -3 2 | | 7 1 3 | so x=2, y=3, z=1. Once you write the matrices out, it's pretty quick even for systems of three variables. edit: That looks terrible on mobile lmao the important thing is that x_i = det(A_i)/det(A), where A_i is the coefficient matrix A with the RHS vector b substituted into the ith column, so 2 1 8 5 -3 3 7 1 20 is the A_3 for z because z=x_3, so we substitute the b=[8,3,20]^T into A in the third column.

Yes, use Gauss Jordan elimination, then Cramer's Rule.

Please do, don't expect a teacher to teach like this video.

Isn't this Cramer's rule?

@@xeriop39274 Seems like it is yeah, but I never learned it

Interesting. Can you give an example with two unknowns where it doesn't work?

Note sure what that is.

@@tecmath crammers rule was it

Nah you weren't. Close though.

@@tecmath nooo i wish i was. Oh well 🤣 nice video! really good tip from you!! i watch ur vids every now and then when they pop up in the feed

Close.

Nah, sorry I was