Question 549251
(1) x + y + z = 5
(2) xy + yz + zx = 3
<br>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)<sup>2</sup> - 2(2), we get:
<br>(3) x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 19. 
<br>This equation is  equation of the sphere whose radius is &radic;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:
<br>(1) x + 2y = 5
(2) xy + y<sup>2</sup> + xy = 3
<br>This can be solved for x and y.
<br>(1) x = 5 - 2y
(2) y<sup>2</sup> + 2xy = 3
<br>By substituting x from (1) in (2), we will end up with:
<br>y<sup>2</sup> + 2(5 - 2y)y = 3, which simplifies to:
3y<sup>2</sup> - 10y + 3 = 0
<br>Using quadratic formula we can get the values for y to be:
y<sub>1</sub> = 1/3, and y<sub>2</sub> = 3.
<br>The first value of y, y<sub>1</sub>, yields x to be:
x = 5 - 2(1/3) = 13/3
<br>The second value of y, y<sub>2</sub>, yields x to be:
x = 5 - 2(3) = -1.
<br>Hence, the minimum value of x is -1 and the maximum value of x is 13/3.
<br>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.