SOLUTION: If x+y+z=5 and xy+yz+zx=3,then least and largest value of x are ?

Question 549251: If x+y+z=5 and xy+yz+zx=3,then least and largest value of x are ?
Answer by karaoz(32) (Show Source): You can put this solution on YOUR website!
(1) x + y + z = 5
(2) xy + yz + zx = 3

I do not see how this can be solved with elementary algebra unless we use some insights from geometry.
Namely, equation (1) is an equation of a plane in 3D space.
When two equations are combined through (1)² - 2(2), we get:

(3) x² + y² + z² = 19.

This equation is equation of the sphere whose radius is √19.
Radius is less than 5, which means that the intersection of the plane (1) and the sphere (3) is a circle.
The intersecting circle is symmetric with respect to the planes x = y, y = z and x = z.
(All these planes will cut the circle in two halves through its diameter).
So, we can "see" that both, minimum and maximum value of x on the intersecting circle will occur when y = z.
Using this insight we can then substitute y = z into both of the original equations to get:

(1) x + 2y = 5
(2) xy + y² + xy = 3

This can be solved for x and y.

(1) x = 5 - 2y
(2) y² + 2xy = 3

By substituting x from (1) in (2), we will end up with:

y² + 2(5 - 2y)y = 3, which simplifies to:
3y² - 10y + 3 = 0

Using quadratic formula we can get the values for y to be:
y₁ = 1/3, and y₂ = 3.

The first value of y, y₁, yields x to be:
x = 5 - 2(1/3) = 13/3

The second value of y, y₂, yields x to be:
x = 5 - 2(3) = -1.

Hence, the minimum value of x is -1 and the maximum value of x is 13/3.

Normally, this type of problem can be solved as a constrained optimization problem, which uses Lagrange multipliers and calculus to get the extreme points. I am not sure if there is "an easy" algebraic solution to the problem.

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