`x^(2) -8x + 15=0, y^(2)-3y +2=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 equations \(x^2 - 8x + 15 = 0\) and \(y^2 - 3y + 2 = 0\), we will follow these steps:

### Step 1: Solve the first equation \(x^2 - 8x + 15 = 0\)

We can factor this quadratic equation. We need to find two numbers that multiply to \(15\) (the constant term) and add up to \(-8\) (the coefficient of \(x\)).

The numbers that satisfy these conditions are \(-5\) and \(-3\).

So, we can write:
\[ x^2 - 8x + 15 = (x - 5)(x - 3) = 0 \]

### Step 2: Set each factor to zero

Now we set each factor equal to zero:
1. \(x - 5 = 0 \Rightarrow x = 5\)
2. \(x - 3 = 0 \Rightarrow x = 3\)

Thus, the solutions for \(x\) are:
\[ x = 5 \quad \text{and} \quad x = 3 \]

### Step 3: Solve the second equation \(y^2 - 3y + 2 = 0\)

Similarly, we will factor this quadratic equation. We need to find two numbers that multiply to \(2\) and add up to \(-3\).

The numbers that satisfy these conditions are \(-2\) and \(-1\).

So, we can write:
\[ y^2 - 3y + 2 = (y - 2)(y - 1) = 0 \]

### Step 4: Set each factor to zero

Now we set each factor equal to zero:
1. \(y - 2 = 0 \Rightarrow y = 2\)
2. \(y - 1 = 0 \Rightarrow y = 1\)

Thus, the solutions for \(y\) are:
\[ y = 2 \quad \text{and} \quad y = 1 \]

### Step 5: Compare the values of \(x\) and \(y\)

Now we have:
- \(x\) can take values \(5\) or \(3\)
- \(y\) can take values \(2\) or \(1\)

To compare:
- The maximum value of \(x\) is \(5\), which is greater than both values of \(y\) (2 and 1).
- The minimum value of \(x\) is \(3\), which is greater than both values of \(y\) (2 and 1).

### Conclusion

Since both values of \(x\) are greater than both values of \(y\), we conclude that:
\[ x > y \]