﻿ If x^2+xy+y^2=39, y^2+yz+z^2=49, z^2+zx+x^2=19, find all soltion sets for of x, y , z

#### Question

If x^2+xy+y^2=39, y^2+yz+z^2=49, z^2+zx+x^2=19, find all soltion sets for of x, y , z

Collected in the board: System of equations

Steven Zheng posted 2 weeks ago

Multiply the first equation x^2+xy+y^2=39 by x-y, using the difference of cubes identity x^3-y^3 = (x-y)(x^2+xy+y^2), then the first equation is transformed

x^3-y^3 = 39(x-y)
(1)

Similarly, multiply the second equation by y-z, then we get

y^3-z^3 = 49(y-z)
(2)

Multiply the third equation by z-x

z^3-x^3 = 19(z-x)
(3)

Addition of the three equations gives

39(x-y)+49(y-z)+19(z-x)=0

Expand the brackets and simplify

2x+y-3z=0, then

z^2+z

or

y = 3z-2x
(4)

Substitute (4) to the first two equations x^2+xy+y^2=39, y^2+yz+z^2=49

x^2+x(3z-2x)+(3z-2x)^2=39

Simplifying gives the equation

x^2-3zx+3z^2 = 13

Now we have the system of equations

\begin{cases} x^2-3zx+3z^2 = 13 \\ z^2+zx+x^2=19 \end{cases}
(5)

To solve the system of equations, multiply the first equation by 19 and the second one by 13. Then the constant is eliminated and we get the following equation.

19(x^2-3zx+3z^2) = 13(z^2+zx+x^2)

Simplifying the equation gives

6x^2-70zx+44z^2=0

Cancel the common factor

3x^2-35zx+22z^2=0
(6)

Factor the expression

(3x-2z)(x-11z)=0

Then we get

x=\dfrac{2}{3}z
(7)

or

x = 11z
(8)

Substitute x=\dfrac{2}{3}z to the third equation z^2+zx+x^2=19, then

z^2+\dfrac{2}{3}z^2+\dfrac{4}{9}z^2 = 19

19z^2 = 19\cdot 9

z=\pm3
(9)

Substitute to (7), we get

x=\pm2
(10)

Substitute (9) and (10) to (4), we get

y = 3z-2x = \pm5
(11)

So we get two solution set \{2,3,5\} and \{-2,-3,-5\}

Substitute x=11z to the third equation z^2+zx+x^2=19, then

z^2+11z^2+121z^2 = 19

133z^2 = 19

Cancel the common factor 19, we get

z^2 = \dfrac{1}{7}

Solve for z

z=\pm\dfrac{\sqrt{7} }{7}
(12)

Substite to (8) x = 11z , then we get

x =\pm\dfrac{11\sqrt{7} }{7}
(13)

Substitute both (12) and (13) to (4), we get

y = 3z-2x

= 3\Big( \pm\dfrac{\sqrt{7} }{7}\Big) -2\Big( \pm\dfrac{11\sqrt{7} }{7} \Big)

=\pm\dfrac{19\sqrt{7} }{7}

Therefore, we get another two solution set \{ \dfrac{11\sqrt{7} }{7}, -\dfrac{19\sqrt{7} }{7}, \dfrac{\sqrt{7} }{7} \} and \{ -\dfrac{11\sqrt{7} }{7}, \dfrac{19\sqrt{7} }{7}, -\dfrac{\sqrt{7} }{7} \}

Overall, there are 4 solution sets for the system of equations

Steven Zheng posted 2 weeks ago

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