In each of the following questions, two equations are given. You have to solve them
`"I. "sqrt(961)x+1234=1486" II. "sqrt(1024)y-1196=1001`

Options

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

Answer

To solve the given equations step by step, let's start with the first equation:

### Step 1: Solve the first equation
**Equation I:**
\[ \sqrt{961}x + 1234 = 1486 \]

1. Calculate \(\sqrt{961}\):
\[ \sqrt{961} = 31 \]

2. Substitute this value back into the equation:
\[ 31x + 1234 = 1486 \]

3. Isolate \(31x\) by subtracting 1234 from both sides:
\[ 31x = 1486 - 1234 \]
\[ 31x = 252 \]

4. Divide both sides by 31 to solve for \(x\):
\[ x = \frac{252}{31} \]
\[ x = 8.129 (approximately) \]

### Step 2: Solve the second equation
**Equation II:**
\[ \sqrt{1024}y - 1196 = 1001 \]

1. Calculate \(\sqrt{1024}\):
\[ \sqrt{1024} = 32 \]

2. Substitute this value back into the equation:
\[ 32y - 1196 = 1001 \]

3. Isolate \(32y\) by adding 1196 to both sides:
\[ 32y = 1001 + 1196 \]
\[ 32y = 2197 \]

4. Divide both sides by 32 to solve for \(y\):
\[ y = \frac{2197}{32} \]
\[ y = 68.65625 (approximately) \]

### Step 3: Compare \(x\) and \(y\)
Now we have the approximate values:
- \(x \approx 8.129\)
- \(y \approx 68.65625\)

Since \(y\) is greater than \(x\), we conclude:
\[ y > x \]

### Final Answer:
The correct option is that \(y\) is greater than \(x\).

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