X = ?, Y = ?, Z = ? Solve This System of Equations | Step-by-Step Tutorial

[zplusacademy]

Xtreme beautiful effort and execution dear sir regarding this type problem ❤️🙏🙏❤️🙏🙏🙏🙏🙏🎄🌎🌎♾️⛄⛄💕💕👍

[RealSlopeDude]

It took me a couple minutes to play around with the numbers to solve it. But what I have a hard time seeing is where you get the idea for the direction to start your formal solution process. Also, at 10 minutes when you solved the cubic equation, couldn't you have then concluded -2, 3, and 5 as solutions for x, y, and z since the initial systems show that x, y, and z are arbitrarily interchangable? Very beautiful video. Thanks.

[guyhoghton]

If we write x, y and z as the roots of a cubic equation in (say) u:
(u - x)(u - y)(u - z) = 0 ... ①
and expand the left hand side we get:
u³ - (x + y + z)u² + (xy + xz + yz)u - xyz = 0.
Then using your evaluations we can substitute for the coefficients of u:
u³ - 6u² - u + 30 = 0.
As shown in the video this factorises as:
(u + 2)(u - 3)(u - 5) = 0.
∴ x, y, z are the 3 roots -2, 3, 5 respectively by inspection of ①.
In other words: once you've solved the cubic, you've finished.

[drpkmath]

Hey my friend. Similar questions only with different answer parts on each equation were solved before haha Great job. Keep rocking my friend!

[susennath]

Excellent

I specially nostralgic in synthetic division method.
When I was read in degree class in mathematics, I loved this method in classical algebra section.
Also found this derived method in linear programming problem also.
Now I am in that time relativity area.

[peterribers]

The first part of the procedure is nice work. The last part too, but suddenly only integers are considered.
Knowing from the beginning that we seek an integer solution: EQUATION #2 IS SUFFICIENT to find the answer! (eq1 only needed to fix the signs)
The possible candidates are very few. Testing systematically: 0, 0, 0? no! 0, 0, 1? No! 0, 1, 1? No! ...and so forth...
very fast leads to -2, 3, 5. (Most of the candidates like 0, 0, 0 are obviously nogo).

[pwmiles]

If  the polynomials are symmetric, like here, I use the Newton-Girard formulae (thanks Presh!) which give Eqs 4 and 5. Also when you solve the cubic the three roots immediately give x, y, and z.

[golddddus]

Excellent job. I like synthetic division to.

[sanjaysurya]

Thanks a bunch .. Keep feeding our appetite with much tough problems 👍 Excellent 👏

[wackojacko]

I love every famous formula..thank you so much for posting these tutorials.

[govindashit]

More calculative problem.
Thanks dear.

[broytingaravsol]

sth can be trickier for that 3x must be less than 6, for x being less than 2, or just the absolute value of x (since there is a second one with a sum of squares for the 3 absolute values with being randomized in a definite interval to be met with), so there are -2~2, at least the 5 choices to try with the third equation, so with x=-2, 152 can be a sum of 3 and 5 with each corresponding cubic power

[math]

very well done, thanks for sharing, happy holidays

[predator]

Fantastic solution👍, thank you teacher 🙏.

[bentels]

Very nice. I actually got pretty far with this one on my own, but dividing xyz by {x, y, z} didn't occur to me...

No fair not including x <= y <= z on the thumbnail, BTW.

[johndepledge]

Solved pretty much by inspection. I noticed that 4 + 9 + 25 = 38 & that -8 + 27 + 125 = 144. Answers follow. Thank you for the problem.

[SenorQuichotte]

solved by inspection 5 3 -2. you need to specify integers for real and complex numbers there are many solutions. do question is incomplete

[philipkudrna]

Before watching: I didn‘t even dare to attempt the proper long way, but simply tested out some combinations. It turns out that 3+5-2 happens to be 6. (-2)^2+3^2+5^2 or 4+9+25=38. -8+27+125 happens to be 144. So x=-2; y=3 and z=5 definitely are a set of solutions. But are they the only solutions? Probably not. Now I watch the video… :-)
17 minutes and some cumbersome calculations later: The solution turns out to be: x=-2; y=3 and z=5! Hm…

[pierreneau]

When you have S=x+y+z; Q=xy+xz+yz and P=xyz, you know that x, y, z are roots of X^3-SX^2+QX-P=0.

[eleishar]

That's so easy I made it in like 30 seconds.