In the following questions two equations I. and II. are given. You have to solve both the equations and mark the appropriate options
I. ` ( x - 2) ^(2) = x - 2 `
II. ` 9 y ^(2) - 36 y + 35 = 0 `

Options

A. ` x lt y `
B. ` x gt y`
C. ` x le y `
D. x = y or no relation

Answer

To solve the given equations step by step, we will start with Equation I and then proceed to Equation II.

### Step 1: Solve Equation I
The first equation is:
\[ (x - 2)^2 = x - 2 \]

**Hint for Step 1:** Start by expanding the left side of the equation.

Expanding the left side:
\[ x^2 - 4x + 4 = x - 2 \]

**Hint for Step 2:** Move all terms to one side of the equation to set it to zero.

Rearranging the equation:
\[ x^2 - 4x + 4 - x + 2 = 0 \]
\[ x^2 - 5x + 6 = 0 \]

**Hint for Step 3:** Factor the quadratic equation.

Factoring the quadratic:
\[ (x - 2)(x - 3) = 0 \]

**Hint for Step 4:** Set each factor to zero to find the values of \(x\).

Setting each factor to zero:
1. \(x - 2 = 0 \Rightarrow x = 2\)
2. \(x - 3 = 0 \Rightarrow x = 3\)

So, the solutions for Equation I are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

### Step 2: Solve Equation II
The second equation is:
\[ 9y^2 - 36y + 35 = 0 \]

**Hint for Step 5:** Use the quadratic formula or factor the equation.

To factor, we look for two numbers that multiply to \(9 \times 35 = 315\) and add to \(-36\). The numbers are \(-21\) and \(-15\).

Rewriting the equation:
\[ 9y^2 - 21y - 15y + 35 = 0 \]

**Hint for Step 6:** Group the terms to factor by grouping.

Grouping the terms:
\[ (9y^2 - 21y) + (-15y + 35) = 0 \]
Factoring each group:
\[ 3y(3y - 7) - 5(3y - 7) = 0 \]

**Hint for Step 7:** Factor out the common term.

Factoring out the common term:
\[ (3y - 7)(3y - 5) = 0 \]

**Hint for Step 8:** Set each factor to zero to find the values of \(y\).

Setting each factor to zero:
1. \(3y - 7 = 0 \Rightarrow y = \frac{7}{3}\)
2. \(3y - 5 = 0 \Rightarrow y = \frac{5}{3}\)

So, the solutions for Equation II are:
\[ y = \frac{7}{3} \quad \text{and} \quad y = \frac{5}{3} \]

### Summary of Solutions
- From Equation I, we have \(x = 2\) and \(x = 3\).
- From Equation II, we have \(y = \frac{7}{3}\) and \(y = \frac{5}{3}\).

### Final Comparison
Now, we can compare the values of \(x\) and \(y\):
- \(x = 2\) (which is approximately \(2.00\)) and \(y = \frac{5}{3} \approx 1.67\) ⇒ \(x > y\)
- \(x = 3\) and \(y = \frac{7}{3} \approx 2.33\) ⇒ \(x > y\)

Thus, we can conclude that:
- \(x\) is greater than \(y\) in both cases.