`x^(2)-16=0, y^(2)-9y + 20 = 0`

Options

A. if `x gt y`
B. if `x gt= y`
C. if `x lt y`
D. if `x lt= y`

Answer

To solve the given equations \(x^2 - 16 = 0\) and \(y^2 - 9y + 20 = 0\), we will find the values of \(x\) and \(y\) step by step.

### Step 1: Solve for \(x\)
We start with the equation:
\[ x^2 - 16 = 0 \]
This can be factored as:
\[ x^2 - 4^2 = 0 \]
Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), we rewrite it as:
\[ (x - 4)(x + 4) = 0 \]
Setting each factor to zero gives us:
\[ x - 4 = 0 \quad \text{or} \quad x + 4 = 0 \]
Thus, we find:
\[ x = 4 \quad \text{or} \quad x = -4 \]

### Step 2: Solve for \(y\)
Next, we solve the equation:
\[ y^2 - 9y + 20 = 0 \]
To factor this quadratic, we need two numbers that multiply to \(20\) and add to \(-9\). The numbers \(-4\) and \(-5\) fit this requirement:
\[ y^2 - 4y - 5y + 20 = 0 \]
We can group the terms:
\[ y(y - 4) - 5(y - 4) = 0 \]
Factoring out \((y - 4)\):
\[ (y - 4)(y - 5) = 0 \]
Setting each factor to zero gives us:
\[ y - 4 = 0 \quad \text{or} \quad y - 5 = 0 \]
Thus, we find:
\[ y = 4 \quad \text{or} \quad y = 5 \]

### Step 3: Compare the values of \(x\) and \(y\)
Now we have the values:
- \(x = 4\) or \(x = -4\)
- \(y = 4\) or \(y = 5\)

We will compare the values of \(x\) and \(y\):
1. If \(x = 4\) and \(y = 4\), then \(x = y\).
2. If \(x = 4\) and \(y = 5\), then \(x < y\).
3. If \(x = -4\) and \(y = 4\), then \(x < y\).
4. If \(x = -4\) and \(y = 5\), then \(x < y\).

From all comparisons, we can conclude that:
- In all cases, \(x\) is less than or equal to \(y\).

### Final Conclusion
The correct option is:
- \(x \leq y\)