Thank you for sharing this! It’s a great video, has a clear description, is easy to follow work- throughs, and is a great strategy for time management! Keep it coming!
That’s amazing, thanks!! And I don’t expect teachers to know everything but I’ve seen some doozies. The trick is not to point out things during class, but after.
@@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.
I would appreciate knowing why this works. What I mean is that there has to be some underlying structure to why this works. In other words, how did the person who came up with this solution do so?
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
@@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.
Delighted to learn again. All I can guess is that only a few students who have more maths classes and maths electives may be privileged to learn this. We can forget the issue of" we missed out because we did not pay attention in class".
Also if your teacher doesn't know about determinants... and hence this math "trick" (Cauchy wrote about them circa 1800) your in grade 2, or you math teacher is really the gym teacher, hijacked by the principal, to teach O level math.
Thanks for this trick. My Math teacher taught us simultaneous equations and while she was my favourite math teacher, her solution and method were just way too long for me to remember, you made this so easy for me thank you.
This is based on matrices which are have 4 coefficients --- [a b c d] now you, put the coefficients on the matrices and calculate determinant -- (a*d - b*c) = Det (A) Now you put the expressions to left side [a c x y] next determinant (a*x - c*y) = Det (B) Next to right side [x c y d] calculate (x*d - c*y) = Det (C) And the last step is dived the determinants Det (A) / Det (B) Det (A) / Det (C).
Hello sir, I don't expect you to reply, but what do you do if the y and x positions are changed? as I've noticed that the x and y variables were below each other, what if it was switched? how would you solve it? pls tell
What the hell... I never heard about this way of doing it even though I'm studying to become a teacher and will soon pass my national exams! That's a good way to do it... Though I won't try it at the exams, because you never know with the graders...
First of all 2 simultaneous equations, in two unknowns, isn't "ANY Simultaneous Equation." Simultaneous equations can be in at least any number of unknowns. The technique doesn't even work for all simultaneous equations, even in 2 unknowns. As Dave J. ... another famous Australian... is want to say this vid doesn't pass the "sniff" test. But, it's not about math... is it? ... it's about 'tricks' ... so I don't say the the video should be unappreciated.
th-cam.com/play/PLjbxBzUM6SLlqoF9ixrgXCHcbnj1x9lla.html&si=ykdTJj4wOjy6dfrr
OMG!!! I just can't thank you enough. Definitely the best math channel on yt which delivers authentic and practical tricks!
Thank you for sharing this! It’s a great video, has a clear description, is easy to follow work- throughs, and is a great strategy for time management! Keep it coming!
That’s amazing, thanks!! And I don’t expect teachers to know everything but I’ve seen some doozies. The trick is not to point out things during class, but after.
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.
I would appreciate knowing why this works. What I mean is that there has to be some underlying structure to why this works. In other words, how did the person who came up with this solution do so?
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.
Delighted to learn again. All I can guess is that only a few students who have more maths classes and maths electives may be privileged to learn this. We can forget the issue of" we missed out because we did not pay attention in class".
Also if your teacher doesn't know about determinants... and hence this math "trick" (Cauchy wrote about them circa 1800) your in grade 2, or you math teacher is really the gym teacher, hijacked by the principal, to teach O level math.
Thought the same as you. I teach Algebra 2 and this is just Cramer's Rule presented in a different way.
Thanks for this trick. My Math teacher taught us simultaneous equations and while she was my favourite math teacher, her solution and method were just way too long for me to remember, you made this so easy for me thank you.
This is very helpful. I would love to have a video with simultaneous equations but with decimals, fractions and parentheses, thankyou.
This is Cramer's rule and I can guarantee it's taught in at least one high school: mine.
Where were you in my 7th grade algebra class ? I've learned more easier ways to do math from your videos than Mrs Fitzpatrick entire class
No doubt! If my k-12 teachers were as good as he is I wouldn’t have struggled nearly as much as I did
This is great! My 11 year old picked right up on this and got both examples right! Thanks for sharing
Hahaha! I loved this trick of yours & I can’t wait to use it!
Best math teacher on TH-cam😩💗
Thank you Sarah. Hope you liked it.
Like the lattice method for multiplying multiple digit pairs (e.g. 13*14), this technique works faster and easier for smaller digit values.
my guy is doing gods work
Very Good Teaching. SUPERB I understood
Why does this man explain math so well, tbh I'd say it's hot but math is just hot
Looks like Cramer's Rule.
That's because it is.
THANK YOU SO MUCH 😃
Cheers thats much easier than I remember in school!
Isn't this Cramer's rule?
Yes, it is.
Thanks for the refresher. I forgot how to solve these.
It has been too long since the tenth grade.
Thank u thank u I just can’t thank u enough
Good teacher 👍🏻
Great Trick!! Thanks Mate...
This is so powerful time saving quick and smart method intergers is what is required for one to get this right ❤❤ awesome
This is based on matrices which are have 4 coefficients ---
[a b c d] now you, put the coefficients on the matrices and calculate determinant -- (a*d - b*c) = Det (A)
Now you put the expressions to left side [a c x y]
next determinant (a*x - c*y) = Det (B)
Next to right side [x c y d]
calculate (x*d - c*y) = Det (C)
And the last step is dived the determinants Det (A) / Det (B)
Det (A) / Det (C).
Correction *The matrice [a x c y]
That was GREAT!!!
Thanks a lot😊
Wow man that's powerful I also did not know that I wish I new that earliar
Amazing 👏
WHERE WAS THIS WHEN I GRADUATED HIGH SCHOOL 30 YEARS AGO?!
this is just the normal method we were taught in school except made more efficient
Look up Cramer's rule. This is taught in senior high school in some countries.
I'm gonna teach this to my son when he starts learning algebra.
Please do, don't expect a teacher to teach like this video.
Making both coefficients of first and second equation say x to be equal. Equation #1 minus equation #2 ,only y left in the equations.
best maths teacher :) even better than my teacher shhhh dont tell tho lol ;)
thank you for healing my shit-teacheritis
What an awesome and innovative trick to solve equation. Thanks techmath.
Hello sir, I don't expect you to reply, but what do you do if the y and x positions are changed? as I've noticed that the x and y variables were below each other, what if it was switched? how would you solve it? pls tell
Cramers rule using determinants. Learn this and you can solve with 3 unknowns and even higher unknowns.
How would u know when to plus or minus
Your teachers don't know how to do this? Please! Some of us have used and taught this system, using determinants, for decades.
What about having three variables? Does that work? Here is an example:
2x + y + z = 8
5x - 3y +2z = 3
7x + y + 3z = 20
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.
This a good ❤ step I like it
I wish I had met these videos during the past 2 yrs....😢
so you're basically saying, if I use this method for year 10 level simultaneous equations, it'll work and be more efficient?
What the hell... I never heard about this way of doing it even though I'm studying to become a teacher and will soon pass my national exams! That's a good way to do it... Though I won't try it at the exams, because you never know with the graders...
Isn't this Cramer's rule?
@@xeriop39274 Seems like it is yeah, but I never learned it
This is Cramer's Rule, any teacher knows it
how to do this with square numbers?
Gauß-method is way faster for harder questions
😮😮😅yes its true
What if both are in subtraction
This is literally a standard mwthod taught in Grade 9
👍👍👍👍
7:58 has me confused, i think i need to relearn divition rules
Witchcraft!!!
Is this the crammers thing
Note sure what that is.
@@tecmath crammers rule was it
First of all 2 simultaneous equations, in two unknowns, isn't "ANY Simultaneous Equation." Simultaneous equations can be in at least any number of unknowns. The technique doesn't even work for all simultaneous equations, even in 2 unknowns. As Dave J. ... another famous Australian... is want to say this vid doesn't pass the "sniff" test. But, it's not about math... is it? ... it's about 'tricks' ... so I don't say the the video should be unappreciated.
Interesting. Can you give an example with two unknowns where it doesn't work?
😂😂 this is basic math.... you must have had terrible teachers 😂😂😂
LOV
I guess I’m a little dense You lose me on the second part figuring y When you start let’s substitute this number you lose me
first!!
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
FIRST
Close.
Nah, sorry I was
I think you are doing crammers but you have t made the matrix