(i)` 2x^(2)-17x+21 =0`
(ii)` 5y^(2)-32 + 35 = 0`

Options

A. `x lt y`
B. ` x gt y`
C. x = y or no relation can be established between x and y.
D. ` x le y`

Explanation

To solve the given quadratic equations step by step, let's break down each part of the question.

### Part (i): Solve the equation \(2x^2 - 17x + 21 = 0\)

**Step 1: Identify the coefficients**
- The coefficients are \(a = 2\), \(b = -17\), and \(c = 21\).

**Step 2: Calculate the product \(a \cdot c\)**
- Calculate \(a \cdot c = 2 \cdot 21 = 42\).

**Step 3: Find two numbers that multiply to \(42\) and add to \(-17\)**
- The numbers are \(-14\) and \(-3\) because \(-14 \cdot -3 = 42\) and \(-14 + -3 = -17\).

**Step 4: Rewrite the equation using these numbers**
- Rewrite the equation as:
\[ 2x^2 - 14x - 3x + 21 = 0 \]

**Step 5: Factor by grouping**
- Group the terms:
\[ (2x^2 - 14x) + (-3x + 21) = 0 \]
- Factor out common terms:
\[ 2x(x - 7) - 3(x - 7) = 0 \]
- Factor out \((x - 7)\):
\[ (2x - 3)(x - 7) = 0 \]

**Step 6: Set each factor to zero**
- Solve for \(x\):
\[ 2x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2} \]
\[ x - 7 = 0 \quad \Rightarrow \quad x = 7 \]

### Solutions for Part (i):
- The values of \(x\) are \(x = \frac{3}{2}\) and \(x = 7\).

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### Part (ii): Solve the equation \(5y^2 - 32y + 35 = 0\)

**Step 1: Identify the coefficients**
- The coefficients are \(a = 5\), \(b = -32\), and \(c = 35\).

**Step 2: Calculate the product \(a \cdot c\)**
- Calculate \(a \cdot c = 5 \cdot 35 = 175\).

**Step 3: Find two numbers that multiply to \(175\) and add to \(-32\)**
- The numbers are \(-25\) and \(-7\) because \(-25 \cdot -7 = 175\) and \(-25 + -7 = -32\).

**Step 4: Rewrite the equation using these numbers**
- Rewrite the equation as:
\[ 5y^2 - 25y - 7y + 35 = 0 \]

**Step 5: Factor by grouping**
- Group the terms:
\[ (5y^2 - 25y) + (-7y + 35) = 0 \]
- Factor out common terms:
\[ 5y(y - 5) - 7(y - 5) = 0 \]
- Factor out \((y - 5)\):
\[ (5y - 7)(y - 5) = 0 \]

**Step 6: Set each factor to zero**
- Solve for \(y\):
\[ 5y - 7 = 0 \quad \Rightarrow \quad y = \frac{7}{5} \]
\[ y - 5 = 0 \quad \Rightarrow \quad y = 5 \]

### Solutions for Part (ii):
- The values of \(y\) are \(y = \frac{7}{5}\) and \(y = 5\).

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### Summary of Solutions:
- For part (i): \(x = \frac{3}{2}, 7\)
- For part (ii): \(y = \frac{7}{5}, 5\)

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