Solving a 3 by 3 System of Equations (the most organized way) 

For those saying use matrices and Gaussian elimination.... it's basically the exact same algorithm. All a matrix is is allowing you to be cleaner by not looking at the variables. Matrices also wouldn't work for non linear systems, but not to worry, Gaussian elimination still works, it's just a tad different.

The way I did this, before watching: First, let's deal with just the coefficients. What we have is: 1 -1 -4 | 11 5 2 10 | 11 3 4 -4 | 11 Now the 2 most obvious ways to eliminate one variable from 2 of the equations, are to use the 1st eqn. (E1) to cancel either x or y from both the 2nd & 3rd eqns (E2 & E3). Eliminating y will involve slightly smaller numbers, so we'll go with that. We add 2E1 to E2, and 4E1 to E3: 1 -1 -4 | 11 7 0 2 | 33 7 0 -20 | 55 Now subtract E3 from E2, to eliminate x from it: 1 -1 -4 | 11 0 0 22 | -22 7 0 -20 | 55 resulting immediately in z = -1. Which in turn, when substituted into E3, gives 7x + 20 = 55; x = 35/7 = 5 And now, substituting both of those into E1, 5 - y + 4 = 11; y = -2 End result: (x,y,z) = (5, -2, -1) To check, just plug these values into all 3 original equations: 5 + 2 + 4 = 11 25 - 4 - 10 = 11 15 - 8 + 4 = 11 Check! Fred

You could also reduce 14x+4z=66 down to 7x+2z=33 and make it negitive... it’s not that big of a development I just wanted to point that out

you can use matrice and use methode de gause

I solved it by substitution . By 1) and 3) you have X- Y = 3X + 4Y . Thus Y =-2/5X . By 1) Z = 7/20X - 11/4 . By 2) you find X =5 and deduct Y = -2 and Z = -1 .

Wow that actually worked, I was very skeptical of it at first because I usually end up messing up system of 3 linear equations but this method is one of the best I've encountered. Thank you, I have subscribed...

THANK YOU IT MAKES SO MUCH SENSE NOW YOU ARE AMAZING GOD BLESS YOU

eq 3 minus eq 1 gives 2x +5y=0

Another way to solve this and my prefered method is to choose one of the equations and make one of the variables the subject and then substitute back into the other two equations, so in this example make y the subject in the first equation we have y = x - 4z -11. substitute that into the othert two and simplify we get 7x + 2z = 33 and 7x - 20z = 55, take one from the other and you are left with 22z = -22, so z = -1. sub that back into one of the others and we get 7x - 2 = 33, so x = 5 and the go right back to the top equation we get y = 5 + 4 -11. and y = -2. A few lines and its done.

This is not what im needing to learn right now but I wanted to take this moment to tell you how much you are appreciated for all your Cal stuff. Thanks brother.

you can do it with Cramer's

This method is much simpler than any of my teachers who tried to explain how this works, good job

currently doing the 2 variables, this is going to help :)

This has got to be one of the best solution templates for this type of problem. 3 by 3 used to drive me nuts when trying to solve them algebraically. I could easily do this using determinants as someone else mentioned, but on an exam, if the professor is asked for an algebraic solution process determinants get you zero credit for the problem. You have a very bright future in front of you a teacher my young friend. If you can do it such a way that it makes sense to people that's talent. You should be writing textbooks on this.

i like these easy videos because they teach me new ways of learning simple things. i wish my algebra 2 teacher taught it like this. gaussian elimination makes this way easier, but this was cool too

Good method but can be faster. There are several steps that can be quickly get rid of if you understand what you're doing. But other than that this is still much much quicker than determinants and it also looks better. Great for anyone who hasn't developed their own method and/or is struggling with this.

I found an easier way to solve it. Just subtract the third equation from the first. You get rid of BOTH z and 11. Which leaves only x and y, which makes it very easy to find y in terms of x. (y = -2/5 * x). Then you just go back and substitute that for y into the first two equations. You get two equations with only x, z, and 11, which is a system of equations that is relatively easy to solve. E.g. 11=11 so you just set them equal to one another. Or solve both equations for z, and then z=z so set those equal to each other to find x.

For non integer coefficients we use product istead of LCM In some countries this method is known as opposite coefficients

Eliminate Y. Multiply (1) by 2 then add to (2). Multiply (1) by 4 then add to (3). That leaves two equations in two variables (X and Z)...a breeze from there.

I would have used Cramer's rule to solve this.

We can solve it easily X-y-4z= 11 X = 11+y+4z Then substitute x for 2nd and 3rd equation 7y+30z=-44 7y+8z=-22 Subtract these two 22z =-22 Z =-1 Substitute it in one equation 7y-8=-22 7y=-14 Y=-2 Then substitute y and z X= 11-2-4 X=5

thank you so much.

1) 03:05 But you didn't have to multiply the third equation by (-1). You could just leave it as it is: 3x + 4y - 4z = 11, and add it to the first equation (4x - 4y - 16z = 44) to get 7x - 20z = 55. System of equations: 14x + 4z = 66 and 7x - 20z = 55 gives the solutions x = 5, z = (-1) - the same as you get in 7:57. 2) 04:50 I really wonder why you didn't just divide the first equation by (-2) to get smaller numbers to work on... Let;s see: 14x + 4z = 66 | :(-2) 7x + 24z = 11 (-7)x - 2z = (-33) 7x + 24z = 11 Added together, the both equation give us: 22z = (-22) z = (-1) And, the value z = (-1) we put into the second equation: 7x + 24z = 11 7x + 24*(-1) = 11 7x - 24 = 11 7x = 11 + 24 7x = 35 x = 35/7 x= 5 Anyway, thank you so much for the great work you do!

I tried to do that before watching the video, and the results was 77/7=11, I'm a genius.

Hello #bprp, in terms of zeta(x) (if there is an expression that contains zeta(x)), what is the derivative and integral of zeta(x)?

Why is he holding a thermal detonator grenade from Star Wars.

it seems to me that -1 * third equation and the joining of 2nd and 3rd is an unnecessary step.

use rref on your calc

It seemed easier in high school

6:21 why not just divide 14x+4z=66 by 2?

I just used substitution and got the answer within 3 mins, why not do it that way?

How about determinan

surely it would be easier to just set the two equations with -4z in them equal at the start..?

Good thing I'm an EE, we got calculators for that lol

Hey thanks for this video, I was trying to help my son understand solving 3x3 better, and this really helped us both!

See 10:18 video part ha ha ha most liked phrase of mathematics How much I learn from these videos are inderminant form( 1÷0 =infinite)

Matrices!

x=5; y=-2; z=-1?

I solved it in seven minutes with matricies.

Matrice, montante algorithm.

I prefer matrices.

7X and 14X In order to cancel the X you dont have to multiple the 7X by nehetive 2. Multiply 7X by positive 2 is good too. Just dont forget to substrac them from each other. 14X - 14X And walla! The X will be cancled as well.

Use matlab

Haked by calculater EQN. mode.55555554

Using gaussian elimination you couldve done it easier and faster

With a matrix it becomes so much easier

And here I am trying to understand what LCM is....😢

Use a matrix, done in 1 minute.

Cutie.. Teacher..

You are speaking wrong 4,4,10 LCM IS 40 not 20😂😂😂