In each of these questions, two equations I. and II. are give. you have to solve both the equations and give answer
I. ` ( x + 15) ^(2) = ( y + 19) ^(2)`
II. ` x^(2) - y^(2) = 112 `

Options

A. ` x gt y `
B. ` x lt y `
C. ` x ge y `
D. ` x le y`

Answer

To solve the given equations step by step, we have:

### Given Equations:
1. \( (x + 15)^2 = (y + 19)^2 \)
2. \( x^2 - y^2 = 112 \)

### Step 1: Solve the first equation
The first equation can be simplified using the property of squares:

\[ (x + 15)^2 = (y + 19)^2 \]

This implies:

\[ x + 15 = y + 19 \quad \text{or} \quad x + 15 = -(y + 19) \]

**Case 1:**
\[ x + 15 = y + 19 \]
Rearranging gives:
\[ x - y = 4 \quad \text{(Equation A)} \]

**Case 2:**
\[ x + 15 = -y - 19 \]
Rearranging gives:
\[ x + y = -34 \quad \text{(Equation B)} \]

### Step 2: Solve the second equation
The second equation is:

\[ x^2 - y^2 = 112 \]

This can be factored as:

\[ (x - y)(x + y) = 112 \]

### Step 3: Substitute Equation A into the second equation
Using Equation A \( (x - y = 4) \):

\[ 4(x + y) = 112 \]

Solving for \( x + y \):

\[ x + y = \frac{112}{4} = 28 \quad \text{(Equation C)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( x - y = 4 \) (Equation A)
2. \( x + y = 28 \) (Equation C)

We can add these two equations:

\[ (x - y) + (x + y) = 4 + 28 \]

This simplifies to:

\[ 2x = 32 \]

So,

\[ x = \frac{32}{2} = 16 \]

### Step 5: Find the value of y
Substituting \( x = 16 \) back into Equation A:

\[ 16 - y = 4 \]

Solving for \( y \):

\[ y = 16 - 4 = 12 \]

### Final Values
Thus, we have:
- \( x = 16 \)
- \( y = 12 \)

### Conclusion
Now we can state the relationship between \( x \) and \( y \):
- Since \( 16 > 12 \), we conclude that \( x > y \).