In each of the following questions, two equations are given. You have to solve them
`"I. "x^2-6x+9=0" II. "y^2+2y-3=0`

Options

A. `xgty`
B. `xgey`
C. `xlty`
D. `xley`

Answer

To solve the given equations step by step, we will analyze each equation separately.

### Step 1: Solve the first equation \(x^2 - 6x + 9 = 0\)

1. **Identify the equation**: The first equation is \(x^2 - 6x + 9 = 0\).
2. **Factor the equation**: We can factor this quadratic equation. Notice that it can be written as:
\[ (x - 3)(x - 3) = 0 \]
or simply:
\[ (x - 3)^2 = 0 \]
3. **Find the roots**: Setting the factor equal to zero gives:
\[ x - 3 = 0 \implies x = 3 \]
Since it is a perfect square, the root \(x = 3\) has a multiplicity of 2. Thus, the solution for \(x\) is:
\[ x = 3 \]

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

1. **Identify the equation**: The second equation is \(y^2 + 2y - 3 = 0\).
2. **Factor the equation**: We can factor this quadratic equation as follows:
\[ (y + 3)(y - 1) = 0 \]
3. **Find the roots**: Setting each factor to zero gives:
\[ y + 3 = 0 \implies y = -3 \]
\[ y - 1 = 0 \implies y = 1 \]
Thus, the solutions for \(y\) are:
\[ y = -3 \quad \text{and} \quad y = 1 \]

### Step 3: Compare the values of \(x\) and \(y\)

1. **Values obtained**: From the first equation, we have \(x = 3\). From the second equation, we have two values for \(y\): \(y = -3\) and \(y = 1\).
2. **Comparison**:
- Compare \(x = 3\) with \(y = -3\):
\[ 3 > -3 \]
- Compare \(x = 3\) with \(y = 1\):
\[ 3 > 1 \]

### Conclusion

Since \(x = 3\) is greater than both values of \(y\) (which are \(-3\) and \(1\)), we conclude that:
\[ x > y \]

### Final Answer

The correct option is:
**x greater than y**.

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