In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
I `x^2-7x+10=0`
II `y^2-12y+35=0`

Options

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

Answer

To solve the given equations step by step, we will start with each quadratic equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ x^2 - 7x + 10 = 0 \]

To solve this quadratic equation, we can factor it. We need to find two numbers that multiply to \(10\) (the constant term) and add up to \(-7\) (the coefficient of \(x\)).

The numbers that satisfy this are \(-5\) and \(-2\).

So we can rewrite the equation as:
\[ (x - 5)(x - 2) = 0 \]

Now, we set each factor to zero:
1. \( x - 5 = 0 \) → \( x = 5 \)
2. \( x - 2 = 0 \) → \( x = 2 \)

Thus, the solutions for \(x\) are:
\[ x = 5 \quad \text{and} \quad x = 2 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 - 12y + 35 = 0 \]

Again, we will factor this quadratic equation. We need to find two numbers that multiply to \(35\) and add up to \(-12\).

The numbers that satisfy this are \(-7\) and \(-5\).

So we can rewrite the equation as:
\[ (y - 7)(y - 5) = 0 \]

Now, we set each factor to zero:
1. \( y - 7 = 0 \) → \( y = 7 \)
2. \( y - 5 = 0 \) → \( y = 5 \)

Thus, the solutions for \(y\) are:
\[ y = 7 \quad \text{and} \quad y = 5 \]

### Summary of Solutions
The solutions to the equations are:
- For Equation I: \( x = 2 \) and \( x = 5 \)
- For Equation II: \( y = 5 \) and \( y = 7 \)

### Final Relation
Now we can compare the values of \(x\) and \(y\):
- If \(x = 2\), then \(y = 5\) (which gives \(y > x\))
- If \(x = 5\), then \(y = 7\) (which also gives \(y > x\))

Thus, the relation is:
\[ x \leq y \]

### Conclusion
The correct answer is:
\[ x \leq y \]

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