Simultaneous Equations where I let $x + y = 5$

Question

I have the following system of equations $$ x + y + z = 8, \quad xy + yz + xz = 9, \quad xyz = -18$$

I have been given a substitution $x + y = 5$. But I am not sure why or where I can infer this from? Could someone suggest a starting point?

Answer 1

By Vieta's formulae, if $r_1, r_2, r_3$ are the roots of a cubic polynomial $t^3+at^2+bt+c$, then $$r_1+r_2+r_3 = -a, \qquad r_1r_2+r_2r_3+r_3r_1 = b, \qquad \text{and} \qquad r_1r_2r_3 = -c.$$ Therefore, in your example, $x,y,z$ are roots of the equation $$t^3-8t^2+9t+18 = 0 \iff (t-3)(t^2-5t-6) = 0 \\ \iff (t+1)(t-3)(t-6) = 0,$$ i.e. $x,y,z = -1,3,6$ in some order.

Answer 2

From $$xyz = -18$$ $$xy = \frac{-18}{z}$$ and $$x+y=8-z$$

Substituting this in the second equation: $$xy+z(y+x)=9$$ $$\frac{-18}{z}+z(8-z)=9$$ This forms a cubic which bringing into the standard form is: $${z}^{3}-8{z}^{2}+9z+18=0$$ which on solving gives: $$(z-6)(z-3)(z+1)=0$$ $$\therefore z = 6, 3, -1$$

If $z=6$ then, $$x+y=2$$ and $$y=2-x$$ Putting this in the second equation: $$x(2-x)+6(2-x)+6x=9$$ Solves to $$x=-1,3$$

If $x=-1$ then $y=3$ If $x=3$ then $y=-1$

Similarly on solving for the other values of $z$ for $x$ and $y$ we get that: $$x=-1,3,6$$ and $$y=3,-1,6$$

The fact that $x+y=5$ which was mentioned in the question was one of the solutions of z and hence was only partly mentioned.