In each of the following questions two equations are given You have to solve them and
`I 2x^(2)=23 x-63 " " II 2y (y-4)=y`

Options

A. A) If x`gt `y
B. B) If `xgt= y`
C. C) If `x lt y`
D. D) If `x lt= y`

Answer

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

### Step 1: Solve the first equation \(2x^2 = 23x - 63\)

1. Rearrange the equation to set it to zero:
\[ 2x^2 - 23x + 63 = 0 \]

### Step 2: Factor the quadratic equation

2. We need to factor \(2x^2 - 23x + 63\). We look for two numbers that multiply to \(2 \times 63 = 126\) and add to \(-23\). The numbers are \(-14\) and \(-9\).
\[ 2x^2 - 14x - 9x + 63 = 0 \]

3. Group the terms:
\[ (2x^2 - 14x) + (-9x + 63) = 0 \]

4. Factor by grouping:
\[ 2x(x - 7) - 9(x - 7) = 0 \]

5. Factor out the common term:
\[ (2x - 9)(x - 7) = 0 \]

### Step 3: Solve for \(x\)

6. Set each factor to zero:
\[ 2x - 9 = 0 \quad \text{or} \quad x - 7 = 0 \]

7. Solve for \(x\):
- From \(2x - 9 = 0\):
\[ 2x = 9 \implies x = \frac{9}{2} = 4.5 \]
- From \(x - 7 = 0\):
\[ x = 7 \]

### Step 4: Solve the second equation \(2y(y - 4) = y\)

8. Rearrange the equation:
\[ 2y^2 - 8y = y \]

9. Move all terms to one side:
\[ 2y^2 - 9y = 0 \]

### Step 5: Factor the second equation

10. Factor out \(y\):
\[ y(2y - 9) = 0 \]

### Step 6: Solve for \(y\)

11. Set each factor to zero:
\[ y = 0 \quad \text{or} \quad 2y - 9 = 0 \]

12. Solve for \(y\):
- From \(y = 0\):
\[ y = 0 \]
- From \(2y - 9 = 0\):
\[ 2y = 9 \implies y = \frac{9}{2} = 4.5 \]

### Step 7: Compare values of \(x\) and \(y\)

13. We have:
- \(x = 4.5\) or \(x = 7\)
- \(y = 0\) or \(y = 4.5\)

14. Comparing the values:
- The larger value of \(x\) is \(7\) and the larger value of \(y\) is \(4.5\).
- Therefore, \(x\) is greater than \(y\).

### Conclusion

The solution shows that \(x\) is greater than \(y\) when comparing the maximum values obtained.

---