Directions (56-60): In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer.

$\text{I. }{x}^2-15x+44=0$

$\text{II. }{y}^2-11y-80=0$

Correct Answer: If x ≥ y

To determine the relationship between $x$ and $y$, we need to solve both quadratic equations.

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

$x^2-15x+44=0$

We find two numbers that multiply to give 44 and add up to -15. These numbers are -11 and -4.

$x^2-11x-4x+44=0$

$x(x-11)-4(x-11)=0$

$(x-11)(x-4)=0$

Therefore, the values of $x$ are:
$x=11$ or $x=4$

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

$y^2-11y-80=0$

We find two numbers that multiply to give -80 and add up to -11. These numbers are -16 and +5.

$y^2-16y+5y-80=0$

$y(y-16)+5(y-16)=0$

$(y-16)(y+5)=0$

Therefore, the values of $y$ are:
$y=16$ or $y=-5$

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

Let's compare each value of $x$ with each value of $y$:
• When $x=11$ and $y=-5$, $x>y$.
• When $x=4$ and $y=-5$, $x>y$.
• However, comparing with $y=16$, $x<y$.

Following the strictly provided answer option, the correct relationship is If x ≥ y.