Solving system of equations with three unknowns, but equations are missing components?

Question

The equations are:

1. $ y * z = 8$

2. $x + z = 8$

3. $x + y = 6$

Solving for these gets me to:

$z = \frac{16}{7}, y = \frac{7}{2}$ and $x = 8 - \frac{16}{7}$.

This is incorrect however. Putting it into a solver yields no solution.

Why can't this be solved?

Answer 1

$$yz=8\tag{1}$$

$$x+z=8 \implies z=8-x\tag{2}$$

$$x+y=6 \implies y = 6-x\tag{3}$$

substitute $(2)$ and $(3)$ into $(1)$, we get

$$(6-x)(8-x)=8$$

which is a quadratic equation, now you can compute the discriminant to check if there is such real $x$.

• If the discriminant is positive, there are two real solutions for $x$.
• If the dscriminant is zero, there is one real solution for $x$.
• If the dscriminant is negative, there is no solution for $x$.

Answer 2

$z = 8 - x$ and $y = 6 - x$, then $$yz = (6-x)(8-x) = 8$$ which is $$x^2 - 14x + 40 = 0$$ The roots of the above equation are: $$x_1 = 10$$ and $$x_2 = 4$$ For $x_1 = 10$, we get $y_1 = 6-x_1 = -4$ and $z_1 = 8-10 = -2$.

For $x_2 = 4$, we get $y_2 = 6-x_2 = 2$ and $z_2 = 8-4 = 4$.

Therefore two solutions:

$x_1 = 10$, $y_1 = -4$, and $z_1 = -2$.

$x_2 = 4$, $y_2 = 2$ and $z_2 = 4$.

Answer 3

With $(2)$ and $(3)$ one can deduce that $z$ is two greater than $y$ [e.g. $z = y+2$].

With $(1)$ we have $$\begin{align} yz &= 8\\ yz -8&= 0\\ y(y+2) -8&= 0\\ y^2 +2y -8&= 0\\ (y+4)(y-2)&=0\\ \end{align} $$

which gives a solution of $$(x,y,z)\in \lbrace (4, 2, 4), (10, -4, -2)\rbrace $$