In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer
I. ` 5 x ^(2) - 31 x + 30 = 0 `
II. ` 8 y^(2) - 12 y + 4 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y`
C. If ` x lt y `
D. If ` x le y `

Answer

To solve the given equations step by step, we will start with the first equation and then move to the second one.

### Step 1: Solve the first equation
The first equation is:
\[ 5x^2 - 31x + 30 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to \(5 \times 30 = 150\) and add to \(-31\). The numbers are \(-25\) and \(-6\).

We can rewrite the equation as:
\[ 5x^2 - 25x - 6x + 30 = 0 \]

Now, we group the terms:
\[ (5x^2 - 25x) + (-6x + 30) = 0 \]

Factoring out the common terms:
\[ 5x(x - 5) - 6(x - 5) = 0 \]

Now we can factor out \((x - 5)\):
\[ (5x - 6)(x - 5) = 0 \]

Setting each factor to zero gives us:
1. \(5x - 6 = 0 \Rightarrow x = \frac{6}{5}\)
2. \(x - 5 = 0 \Rightarrow x = 5\)

### Step 2: Solve the second equation
The second equation is:
\[ 8y^2 - 12y + 4 = 0 \]

We will factor this quadratic equation. We need to find two numbers that multiply to \(8 \times 4 = 32\) and add to \(-12\). The numbers are \(-8\) and \(-4\).

We can rewrite the equation as:
\[ 8y^2 - 8y - 4y + 4 = 0 \]

Now, we group the terms:
\[ (8y^2 - 8y) + (-4y + 4) = 0 \]

Factoring out the common terms:
\[ 8y(y - 1) - 4(y - 1) = 0 \]

Now we can factor out \((y - 1)\):
\[ (8y - 4)(y - 1) = 0 \]

Setting each factor to zero gives us:
1. \(8y - 4 = 0 \Rightarrow y = \frac{4}{8} = \frac{1}{2}\)
2. \(y - 1 = 0 \Rightarrow y = 1\)

### Step 3: Compare values of \(x\) and \(y\)
Now we have the values:
- For \(x\): \(x = 5\) and \(x = \frac{6}{5}\)
- For \(y\): \(y = 1\) and \(y = \frac{1}{2}\)

Now we will compare these values:
1. \(5 > 1\) and \(5 > \frac{1}{2}\)
2. \(\frac{6}{5} = 1.2 > 1\) and \(\frac{6}{5} > \frac{1}{2}\)

In both cases, we see that \(x\) is greater than \(y\).

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