Solving a Quick and Easy Polynomial System

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[theimmux]

Seeing the top equation you already know the solutions

[yoav]

8 solutions and they all integers, nice!

[robertgerbicz]

To cut the video:
If x+y+z=-12 then just set x=-x, y=-y, z=-z with this the set of equations will be also true but x+y+z=12, so you just need to negate only all variables in the solution.

[antoniosechi]

The system has degree 8, therefore it has a set of 8 complex solutions (x, y, z) including multiplicity.

Square second equation: z^2 * (x+y)^2 = 25*49
Substitute first members of first and third equations:
z^2 * (x+y)^2 = (x^2 + y^2) * (z^2 + 2xy); xy * z^2 = xy * (x^2+y^2)
If x and y are both non zero, then z^2 = x^2+y^2 = 25.
Using second and third equations, we get two systems for x, y:
[x+y = 7; xy = 12]
[x+y= -7; xy = 12]
Solutions (3, 4, 5), (4, 3, 5), (-3, -4, -5), (-4, -3, -5) come easily.

Back to the original system: for x=0, solutions are (0, 5, 7), (0, -5, -7); for y=0, solutions are (5, 0, 7), (-5, 0, -7).
With 8 solutions found, the system is solved.

[frentz]

I love it. "I'm just going to randomly make some adjustments here, let's multiply the second equation by two ..." .. "Suppose you decided after going north for a while, to stop halfway up the next block and cut across the street .. " Wait, what?

[puhbdle]

After obtainig x+y+z=±12 and substituting x+y=-z±12 into (x+y)z=35,
we have z²∓12z+35=(z∓5)(z∓7)=0. ∴z=±5, ±7 (same order).
If z=±5, then x+y=±7 (same order). Using eq. 2xy=49-z² directly leads to xy=12.
x and y are solutions of u²∓7u+12=(u∓3)(u∓4)=0. x=±3, y=±4; x=±4, y=±3.
∴ (x, y, z)=±(3, 4, 5), ±(4, 3, 5).
The case z=±7 is easier and we have (x, y, z)=±(0, 5, 7), ±(5, 0, 7).

[anshumanagrawal]

It's clear from the equations that if that all the variables are negated the equations are still satisfied, so the other case for |(x+y+z)| = 12 didn't need to be solved separately as you'd just get the negatives of the same solutions

[DeJay]

Hearing him say "x plus y plus z quantity squared" reminded me of Michael Penn, funny.

[italixgaming]

My solution (of course way faster):

Then (x+y+z-12)(x+y+z+12)=0.
(2) => 35+z²=z.(x+y+z) => z²-12z+35=0 or z²+12z+35=0 => (z-5)(z-7)=0 or (z+5)(z+7)=0 =>(z²-25)(z²-49)=0
If z²=49 then (3) => xy=0. If for example x=0 then (1) => y²=25 and (2) => yz=35 then we have the solutions: (0, 5, 7), (0, -5, -7), (5, 0, 7) and (-5, 0, -7).
If z²=25 then (3) => xy=12 and (2) => (x+y)=7 or (x+y)=-7 (depending on the sign of z).
If z=5, xy=12 and x+y=7 then x and y are solutions of u²-7u+12=0. These solutions are 3 and 4.
Then we have the following solutions (3, 4, 5) and (4, 3, 5).
If z=-5, xy=12 and x+y=-7 then x and y are solutions of u²+7u+12=0. These solutions are -3 and -4.
Then we have the following solutions: (-3, -4, -5) and (-4, -3, -5).

[carloshuertas]

Another great explanation, SyberMath! I actually solved one solution in my head. Thanks a lot!

[mokouf]

It is very useful to solve these for Calc III.

[JohnRandomness]

I'm not sure that I have all possible solutions, but x² + y² = 25 jumped out at me, so I tried 9 + 16 = 25. I began with x = 3, y = 4. Then z(x + y) = 35 meant z = 5. 49 - 25 = 24 = 2*3*4. So (x, y, z) = (3, 4, 5) is a solution. Also, switching x and y changes nothing -- (4, 3, 5) is a solution. Multiplying the solutions all by (-1) leaves the equations unchanged -- (-3, -4, -5) and (-4, -3, -5) are solutions.

Now, I don't know if there are any other solutions.

[alphalunamare]

Seriously that was a 30 sec mental arithmetic test! :-)

[SamsungJ-kknr]

Very good exercise, is ideal to test.

[threstytorres]

DONT CLICK READ MORE






Since the first equation's sum is a perfect square we can see that the equation is a Pythagorean triple of 3, 4, and 5. X and Y are 3 and 4. to find Z factor Z in the Second Equation from XZ + YZ to
(X+Y)Z since X and Y are 3 and 4, therefore, X+Y=7 and perform 7Z=35. Which makes Z has a value of 5

[Qermaq]

As usual, your approach is more sensible and logical than mine. It's almost as if you've seen the problem before me! :D

[mxsjncv]

If we do not see this nice way, we can take z = 35 / (x+y) and go from there ... 2xy = t

[binaryblade]

guess x^2 + y^2 = 25 is a 3, 4, 5 pythagorean triple.
Confirm consistency with other 3 equations
Not the x, y symmetry

[divyanshsrivastava]

x=4
y=3
z=5 and also other many solns

[erikroberts]

That was so weird, I just looked at the problem for not more than a few seconds before activating the video, and I knew right then and there that the solution was: x=3, y=4, and z=5. Of course, I had to watch the rest of the video, to learn what the rest of the solutions were. Interesting how the mind works sometimes.