In questions two equations numbered I and II are given you have to solve both equations and give answer
`I. 3x^2-22x+40=0 II. 5y^2-21y+16=0`

Options

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

Answer

To solve the given equations, we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 - 22x + 40 = 0 \]

To solve this quadratic equation, we will factor it. We need to find two numbers that multiply to \(3 \times 40 = 120\) and add up to \(-22\).

**Finding the factors:**
The factors of \(120\) that add up to \(-22\) are \(-12\) and \(-10\).

Now we can rewrite the equation:
\[ 3x^2 - 12x - 10x + 40 = 0 \]

**Grouping terms:**
Group the terms:
\[ (3x^2 - 12x) + (-10x + 40) = 0 \]

Factor out common terms:
\[ 3x(x - 4) - 10(x - 4) = 0 \]

Now we can factor by grouping:
\[ (3x - 10)(x - 4) = 0 \]

**Finding the values of x:**
Setting each factor to zero gives us:
1. \(3x - 10 = 0 \Rightarrow x = \frac{10}{3} \approx 3.33\)
2. \(x - 4 = 0 \Rightarrow x = 4\)

### Step 2: Solve Equation II
The second equation is:
\[ 5y^2 - 21y + 16 = 0 \]

Again, we will factor this quadratic equation. We need to find two numbers that multiply to \(5 \times 16 = 80\) and add up to \(-21\).

**Finding the factors:**
The factors of \(80\) that add up to \(-21\) are \(-16\) and \(-5\).

Now we can rewrite the equation:
\[ 5y^2 - 16y - 5y + 16 = 0 \]

**Grouping terms:**
Group the terms:
\[ (5y^2 - 16y) + (-5y + 16) = 0 \]

Factor out common terms:
\[ y(5y - 16) - 1(5y - 16) = 0 \]

Now we can factor by grouping:
\[ (5y - 16)(y - 1) = 0 \]

**Finding the values of y:**
Setting each factor to zero gives us:
1. \(5y - 16 = 0 \Rightarrow y = \frac{16}{5} = 3.2\)
2. \(y - 1 = 0 \Rightarrow y = 1\)

### Step 3: Compare Values of x and y
We have the following values:
- For \(x\): \(\frac{10}{3} \approx 3.33\) and \(4\)
- For \(y\): \(\frac{16}{5} = 3.2\) and \(1\)

Now, we will compare \(x\) and \(y\):
1. \(3.33\) vs \(1\): \(3.33 > 1\)
2. \(3.33\) vs \(3.2\): \(3.33 > 3.2\)
3. \(4\) vs \(1\): \(4 > 1\)
4. \(4\) vs \(3.2\): \(4 > 3.2\)

In all cases, \(x\) is greater than \(y\).

### Final Answer
Thus, the relation between \(x\) and \(y\) is:
\[ x > y \]