What Are the Least and Greatest Values of x?

Assume that $x$, $y$, and $z$ are real numbers, with no additional restrictions. The two given equations determine the sum and product of $y$ and $z$ once you choose $x$. Requiring $y$ and $z$ to be real then gives exactly the range of possible values of $x$.

The least value is $-1$, and the greatest is $\frac{13}{3}$. Both are actually attained, not merely approached.

Express the sum and product of y and z

From $x+y+z=5$, we have

$$y+z=5-x.$$

Using $xy+xz=x(y+z)$ in the second condition,

$$xy+yz+zx=x(y+z)+yz=3.$$

Therefore,

$$yz=3-x(y+z)=3-x(5-x)=x^2-5x+3.$$

For each proposed value of $x$, the numbers $y$ and $z$ must have sum $5-x$ and product $x^2-5x+3$.

Require y and z to be real

Two numbers with sum $S$ and product $P$ are roots of the quadratic equation

$$t^2-St+P=0.$$

Here, that equation is

$$t^2-(5-x)t+(x^2-5x+3)=0,$$

whose roots are $y$ and $z$. A quadratic with real coefficients has real roots exactly when its discriminant is nonnegative. Thus,

$$\begin{aligned} (5-x)^2-4(x^2-5x+3)&\geq 0,\\ 25-10x+x^2-4x^2+20x-12&\geq 0,\\ 13+10x-3x^2&\geq 0. \end{aligned}$$

Factoring gives

$$-(3x-13)(x+1)\geq 0,$$

so

$$-1\leq x\leq \frac{13}{3}.$$

Every value in this interval works: the quadratic for $y$ and $z$ has real roots there. Values outside it make the discriminant negative, so no real $y$ and $z$ can satisfy the conditions.

Check that both endpoints are attained

At either endpoint the discriminant is zero, so $y=z$.

If $x=-1$, then $y+z=6$ and $yz=9$, giving $y=z=3$. The triple $(-1,3,3)$ satisfies both original equations:

$$-1+3+3=5,\qquad (-1)(3)+(3)(3)+(3)(-1)=3.$$

If $x=\frac{13}{3}$, then $y+z=\frac{2}{3}$ and $yz=\frac{1}{9}$, giving $y=z=\frac{1}{3}$. This triple also satisfies the conditions:

$$\frac{13}{3}+\frac{1}{3}+\frac{1}{3}=5,$$ $$\frac{13}{9}+\frac{13}{9}+\frac{1}{9}=3.$$

So the bounds are included, and the requested least and greatest values are $-1$ and $\frac{13}{3}$.

Common mistakes

• Assuming $x$ must be positive. The question gives no positivity condition. The valid triple $(-1,3,3)$ shows that negative $x$ is allowed.
• Checking only that $x$ is real. A real value of $x$ is not enough; the resulting $y$ and $z$ must also be real. The discriminant condition enforces this.
• Giving bounds without checking attainability. A discriminant inequality gives a closed interval here, but verifying the endpoint triples confirms that both extreme values occur.

If the variables are required to be nonnegative instead, the least value remains $0$: for $x=0$, $y$ and $z$ are the positive roots of $t^2-5t+3=0$. If all three variables must be strictly positive, there is no least value of $x$; its infimum is $0$. The greatest value remains $\frac{13}{3}$ in either case.