To solve this system of equations:

1. $x + y + z = 65$
2. $y^2 = x \cdot z$
3. $2y = (x - 1) + (z - 19)$

Let’s break it down step by step:

Step 1: Simplify the third equation

$2y = x - 1 + z - 19$ $2y = x + z - 20$

Step 2: Substitute $x + z$ from Equation (1)

From $x + y + z = 65$, we know: $x + z = 65 - y$

Substitute $x + z = 65 - y$ into the third equation: $2y = (65 - y) - 20$ $2y = 45 - y$ $3y = 45$ $y = 15$

Step 3: Solve for $x + z$

Substitute $y = 15$ into $x + z = 65 - y$: $x + z = 65 - 15$ $x + z = 50$

Step 4: Use the second equation

The second equation is $y^2 = x \cdot z$. Substitute $y = 15$: $15^2 = x \cdot z$ $225 = x \cdot z$

Step 5: Solve for $x$ and $z$

We now have two equations:

1. $x + z = 50$
2. $x \cdot z = 225$

Let $x$ and $z$ be the roots of the quadratic equation: $t^2 - (x+z)t + x \cdot z = 0$ Substitute $x + z = 50$ and $x \cdot z = 225$: $t^2 - 50t + 225 = 0$

Solve this quadratic equation using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -50$, $c = 225$: $t = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(225)}}{2(1)}$ $t = \frac{50 \pm \sqrt{2500 - 900}}{2}$ $t = \frac{50 \pm \sqrt{1600}}{2}$ $t = \frac{50 \pm 40}{2}$

The two solutions are: $t = \frac{50 + 40}{2} = 45$ $t = \frac{50 - 40}{2} = 5$

Thus, $x = 45$ and $z = 5$, or vice versa.

Final Solution:

• $x = 45$, $y = 15$, $z = 5$
• Alternatively, $x = 5$, $y = 15$, $z = 45$

Would you like a detailed explanation of any step? Here are some related questions:

1. How do you use substitution effectively in solving equations?
2. Can you explain the quadratic formula in more detail?
3. How does the discriminant in the quadratic formula help determine the roots?
4. What is the significance of symmetric equations like $x+z = 50$ in problem-solving?
5. Can you graph these solutions to visualize them?

Tip: Always check your solutions by substituting them back into the original equations to ensure they satisfy all conditions.