In each of the following questions, two equations are given. You have to solve them and give answer(निम्नलिखित प्रत्येक प्रश्न में दो समीकरण दिए गए हैं। आपको उन्हें हल करना है और उत्तर देना है)
I. `34x^(2) + 11x = 3`
II. `51y^(2) -77 y + 12 = 0`

Options

A. `xgty`
B. `xgey`
C. `x lty`
D. `xley`

Answer

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

### Step 1: Solve the first equation
The first equation is:
\[ 34x^2 + 11x - 3 = 0 \]

This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 34 \)
- \( b = 11 \)
- \( c = -3 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) is calculated using the formula:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = 11^2 - 4 \cdot 34 \cdot (-3) \]
\[ D = 121 + 408 \]
\[ D = 529 \]

### Step 3: Find the roots using the quadratic formula
The roots of the equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ x = \frac{-11 \pm \sqrt{529}}{2 \cdot 34} \]
Since \( \sqrt{529} = 23 \):
\[ x = \frac{-11 \pm 23}{68} \]

Calculating the two possible values for \( x \):
1. \( x_1 = \frac{-11 + 23}{68} = \frac{12}{68} = \frac{3}{17} \)
2. \( x_2 = \frac{-11 - 23}{68} = \frac{-34}{68} = -\frac{1}{2} \)

### Step 4: Solve the second equation
The second equation is:
\[ 51y^2 - 77y + 12 = 0 \]

This is also a quadratic equation where:
- \( a = 51 \)
- \( b = -77 \)
- \( c = 12 \)

### Step 5: Calculate the discriminant for the second equation
\[ D = (-77)^2 - 4 \cdot 51 \cdot 12 \]
\[ D = 5929 - 2448 \]
\[ D = 3481 \]

### Step 6: Find the roots using the quadratic formula for the second equation
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ y = \frac{77 \pm \sqrt{3481}}{2 \cdot 51} \]
Since \( \sqrt{3481} = 59 \):
\[ y = \frac{77 \pm 59}{102} \]

Calculating the two possible values for \( y \):
1. \( y_1 = \frac{77 + 59}{102} = \frac{136}{102} = \frac{68}{51} = \frac{4}{3} \)
2. \( y_2 = \frac{77 - 59}{102} = \frac{18}{102} = \frac{3}{17} \)

### Step 7: Compare the values of \( x \) and \( y \)
From our calculations:
- \( x_1 = \frac{3}{17} \)
- \( x_2 = -\frac{1}{2} \)
- \( y_1 = \frac{4}{3} \)
- \( y_2 = \frac{3}{17} \)

Now, we compare:
- \( x_1 = \frac{3}{17} \) and \( y_2 = \frac{3}{17} \) (they are equal)
- \( x_2 = -\frac{1}{2} \) is less than both \( y_1 \) and \( y_2 \)

### Conclusion
Since \( x_1 = y_2 \), we can conclude that \( x \) is less than or equal to \( y \).

### Final Answer
The answer is \( x \leq y \).

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