document.write( "Question 549251: If x+y+z=5 and xy+yz+zx=3,then least and largest value of x are ? \n" ); document.write( "
Algebra.Com's Answer #357997 by karaoz(32)\"\" \"About 
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(1) x + y + z = 5
\n" ); document.write( "(2) xy + yz + zx = 3
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I do not see how this can be solved with elementary algebra unless we use some insights from geometry.
\n" ); document.write( "Namely, equation (1) is an equation of a plane in 3D space.
\n" ); document.write( "When two equations are combined through (1)2 - 2(2), we get:
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(3) x2 + y2 + z2 = 19.
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This equation is equation of the sphere whose radius is √19.
\n" ); document.write( "Radius is less than 5, which means that the intersection of the plane (1) and the sphere (3) is a circle.
\n" ); document.write( "The intersecting circle is symmetric with respect to the planes x = y, y = z and x = z.
\n" ); document.write( "(All these planes will cut the circle in two halves through its diameter).
\n" ); document.write( "So, we can \"see\" that both, minimum and maximum value of x on the intersecting circle will occur when y = z.
\n" ); document.write( "Using this insight we can then substitute y = z into both of the original equations to get:
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(1) x + 2y = 5
\n" ); document.write( "(2) xy + y2 + xy = 3
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This can be solved for x and y.
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(1) x = 5 - 2y
\n" ); document.write( "(2) y2 + 2xy = 3
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By substituting x from (1) in (2), we will end up with:
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y2 + 2(5 - 2y)y = 3, which simplifies to:
\n" ); document.write( "3y2 - 10y + 3 = 0
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Using quadratic formula we can get the values for y to be:
\n" ); document.write( "y1 = 1/3, and y2 = 3.
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The first value of y, y1, yields x to be:
\n" ); document.write( "x = 5 - 2(1/3) = 13/3
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The second value of y, y2, yields x to be:
\n" ); document.write( "x = 5 - 2(3) = -1.
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Hence, the minimum value of x is -1 and the maximum value of x is 13/3.
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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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