In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer
I. ` 4 x ^(2) - 17 x + 15 = 0 `
II. ` 2 y^(2) - 17 y + 35 = 0 `

Options

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

Answer

To solve the given equations, we will proceed step by step.

### Step 1: Solve Equation I
The first equation is:
\[ 4x^2 - 17x + 15 = 0 \]

This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 4 \)
- \( b = -17 \)
- \( c = 15 \)

To solve this, we can use the factorization method. We need to find two numbers that multiply to \( ac = 4 \times 15 = 60 \) and add up to \( b = -17 \).

#### Finding Factors:
We need two factors of 60 that add up to -17. The factors are -12 and -5:
- \( -12 \times -5 = 60 \)
- \( -12 + -5 = -17 \)

#### Rewriting the Equation:
We can rewrite the equation as:
\[ 4x^2 - 12x - 5x + 15 = 0 \]

Now, we can group the terms:
\[ (4x^2 - 12x) + (-5x + 15) = 0 \]

Factoring by grouping:
\[ 4x(x - 3) - 5(x - 3) = 0 \]

Now, factor out the common term:
\[ (4x - 5)(x - 3) = 0 \]

#### Finding the Roots:
Setting each factor to zero gives us:
1. \( 4x - 5 = 0 \) → \( x = \frac{5}{4} = 1.25 \)
2. \( x - 3 = 0 \) → \( x = 3 \)

So, the solutions for \( x \) are:
\[ x = 1.25 \quad \text{and} \quad x = 3 \]

### Step 2: Solve Equation II
The second equation is:
\[ 2y^2 - 17y + 35 = 0 \]

Again, this is a quadratic equation where:
- \( a = 2 \)
- \( b = -17 \)
- \( c = 35 \)

To solve this, we need to find two numbers that multiply to \( ac = 2 \times 35 = 70 \) and add up to \( b = -17 \).

#### Finding Factors:
The factors of 70 that add up to -17 are -10 and -7:
- \( -10 \times -7 = 70 \)
- \( -10 + -7 = -17 \)

#### Rewriting the Equation:
We can rewrite the equation as:
\[ 2y^2 - 10y - 7y + 35 = 0 \]

Now, we can group the terms:
\[ (2y^2 - 10y) + (-7y + 35) = 0 \]

Factoring by grouping:
\[ 2y(y - 5) - 7(y - 5) = 0 \]

Now, factor out the common term:
\[ (2y - 7)(y - 5) = 0 \]

#### Finding the Roots:
Setting each factor to zero gives us:
1. \( 2y - 7 = 0 \) → \( y = \frac{7}{2} = 3.5 \)
2. \( y - 5 = 0 \) → \( y = 5 \)

So, the solutions for \( y \) are:
\[ y = 3.5 \quad \text{and} \quad y = 5 \]

### Step 3: Compare Values of \( x \) and \( y \)
Now we have the values:
- For \( x \): \( 1.25 \) and \( 3 \)
- For \( y \): \( 3.5 \) and \( 5 \)

We can compare:
- \( 1.25 < 3.5 \)
- \( 1.25 < 5 \)
- \( 3 < 3.5 \)
- \( 3 < 5 \)

### Conclusion
From the comparisons, we can conclude that:
\[ x < y \]