Solving a Nice Polynomial System in Two Ways

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[kabirsethi]

Let x+y+z=K
xK=1
yK=2
zK=6
Add up all three and get k common.
K(x+y+z)=9
But x+y+z=K
Hence, K²=9 and x+y+z is 3. Now substitute xK=1 hence x=1/3 similarly solve for all.
Thank you for the problem enjoyed solving it.

[gdtargetvn]

Add all 3 equations together and we get (x+y+z)^2=9
=> x+y+z=3 or x+y+z=-3
Subsitute those in, and the remaining parts are easy.

[mcwulf]

Easy. z=6x, y=2x so subst in (1) 9x^2 = 1, x= +/- (1/3) etc.
Solved in one line.

[romanvolotov]

I solved it with the second method in my head in like 15 seconds or so, not that hard

[seanfraser]

Add the three equations together and find
(x+y+z)^2 = 9
So S =x+y+z=3 or -3.

If S=3, then 3x = 1, 3y=2, and 3z=6. So x=1/3, y=2/3, and z = 2.
Similarly if S=-3, then x=-1/3, y=-2/3, and z=-2

[titassamanta]

Add them all and u will end up getting (x+y+z)^2=9 whence x+y+z= +- 3. Then u can eliminate.

[GirishManjunathMusic]

Before watching the video:
Given:
1 — x(x + y + z) = 1
2 — y(x + y + z) = 2
3 — z(x + y + z) = 6

Adding 1, 2, and 3:
x(x + y + z) + y(x + y + z) + z(x + y + z) = 1 + 2 + 6
(x + y + z)(x + y + z) = 9
(x + y + z)² = 9
x + y + z = ±3

Substituting into 1, 2, and 3
3x = 1 or –3x = 1
3y = 2 or –3y = 2
3z = 6 or –3z = 6
x = ⅓ or –⅓
y = ⅔ or –⅔
z = 2 or – 2

Cross-checking by back-substitution:
for x + y + z = 3: ⅓ + ⅔ + 2 = 3
for x + y + z = –3: –⅓ –⅔ –2 = –3

Thus x + y + z = ±3
And
x = ±⅓
y = ±⅔
z = ±2.

[cicik]

oka, y you instantly see the ration of x/y/z = 1/2/6, so x+y+z = x+2x+6x = 9x,
9x² = 1
x = -+1/3 s0
y=-+2/3
z = -+2, all with same sign.
check, if x = -1/3
-2*(-3) = 6

[GhostyOcean]

Since the products are all non-zero, then x, y, z, and x+y+z are all non-zero. Multiply equation 1 by 2 to get
2x(x+y+z)=2=y(x+y+z), so y=2x. Similarly with the second and third equations, z=3y=6x.
Then solve like you did.

[devondevon]

1/x= 2/y= 6/z since all = x+ y+ z
1/x= 2/y
hence y= 2x
1/x= 6/z
hence z=6x
hence ( x + 2x + 6x)x= 1 substitute into the first equation "x(x+y+z)=1"
9x^2=1
x^2= 1/9
x= + or - 1/3
since y=2x, then y= 2/3 and -1/4
since z=6x then z = 2 and -2
answer 1/3, 2/3, 2 AND -1/3, -2/3, 2

[samuelmarger]

I use a way similar to the 1st method. The problem was not difficult but fun nonetheless.

[satrajitghosh]

Adding these three equations one gets
EITHER x+y+z = 3,
x = x(x+y+z)/(x+y+z) =1/3
y = y(x+y+z)/(x+y+z) =2/3
z = z(x+y+z)/(x+y+z) =6/3 = 2
OR x+y+z = -3,
x = x(x+y+z)/(x+y+z) = -1/3
y = y(x+y+z)/(x+y+z) = -2/3
z = z(x+y+z)/(x+y+z) = -6/3 = -2
THAT IS THE ANSWER

[mathswan]

adding 3 eqns we have x+y+z=+/-3 so we have x=1/3 y=2/3 and z=2 or x=-1/3 y=-2/3 and z=-2

[misterdubity]

One advantage of the upside down -/+ instead of +/- is that if you're writing this on paper and this occurs in a numerator, the minus part of the usual +/- can get buried in the horizontal line separating the numerator from the denominator. Addendum: curious that the editor put in a strikethrough. I didn't type it that way.

[ALeafOnTheWind]

I did a variation on the first method, though it boils down to the same thing: we can immediately conclude none of x, y, z are 0, so we can divide (1) by x, (2) by y, and (3) by z. Then because the LHS in all 3 are x+y+z, we conclude that 1/x=2/y=6/z. This creates substitutions we can make to get the solutions.

I do like your 2nd method though. It's elegant

[Skank_and_Gutterboy]

Rearrange the 3 equations:
x + y + z = 1/x
x + y + z = 2/y
x + y + z = 6/z

Then it's easy to see that the right-hand sides of the equations are all equal. With some rearrangement:
y = 2x
z = 3y
z = 6x

Substituting into the equation x + y + z = 1/x:
x + 2x + 6x = 1/x
So: 9x^2 = 1 and x = (1/3, -1/3)

Then y and z are easily calculated:
y = 2x = (2/3, -2/3)
z = 3y = (2, -2)

These values are easy to check against the original equations.
So, (x, y, z) = { (1/3, 2/3, 2), (-1/3, -2/3, -2) }

[robertodiasfb]

I thought you were going to use your favourite method: substitution. I was wrong! 😁

[sushmamanik]

I added all the three equation and got the answers

[darklordbgextrachannel]

Too easy, but very very cool and smart. I liked the second method the most

[MrArcan]

I used the first method. The second one is also nice.