Two quadratic equations, labeled (I) and (II), are given below. Solve both equations to determine the relationship between x and y:

$\text{I.}{x}^2-11x+28=0$

$\text{II.}{y}^2+13y+30=0$

The correct option is If x is greater than y (or $x>y$).

To determine the relationship between $x$ and $y$, we need to solve both quadratic equations step-by-step to find their roots.

Step 1: Solve Equation (I) for $x$

The given equation is:

$x^2-11x+28=0$

We need to find two numbers whose product is $28$ and whose sum is $-11$. These numbers are $-7$ and $-4$.

Splitting the middle term:

$x^2-7x-4x+28=0$

$x(x-7)-4(x-7)=0$

$(x-7)(x-4)=0$

So, the values of $x$ are:

$x=7$ or $x=4$

Step 2: Solve Equation (II) for $y$

The given equation is:

$y^2+13y+30=0$

We need to find two numbers whose product is $30$ and whose sum is $13$. These numbers are $10$ and $3$.

Splitting the middle term:

$y^2+10y+3y+30=0$

$y(y+10)+3(y+10)=0$

$(y+10)(y+3)=0$

So, the values of $y$ are:

$y=-10$ or $y=-3$

Step 3: Compare the values of $x$ and $y$

Let's compare every value of $x$ with every value of $y$:

• When $x=7$ and $y=-10$ ⇒ $x>y$
• When $x=7$ and $y=-3$ ⇒ $x>y$
• When $x=4$ and $y=-10$ ⇒ $x>y$
• When $x=4$ and $y=-3$ ⇒ $x>y$

In all cases, every possible value of $x$ (which are positive) is strictly greater than every possible value of $y$ (which are negative).

Therefore, $x>y$ (If x is greater than y).