Directions: Solve both equations and determine the relationship between x and y.

I. $x^2-11x+24=0$
II. $y^2-3y+2=0$

The correct option is If x > y.

To determine the relationship between $x$ and $y$, we solve both quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

The given equation is:

$x^2-11x+24=0$

We factorize the quadratic expression by splitting the middle term $-11x$ into two parts such that their sum is $-11$ and their product is $24$. The numbers are $-8$ and $-3$:

$x^2-8x-3x+24=0$

Factor out common terms from the pairs:

$x(x-8)-3(x-8)=0$

$(x-8)(x-3)=0$

Setting each factor to zero gives:

$x-8=0\Rightarrow x=8$

$x-3=0\Rightarrow x=3$

So, the roots for $x$ are $x=3$ and $x=8$.

Step 2: Solve Equation II for $y$

The given equation is:

$y^2-3y+2=0$

We split the middle term $-3y$ into two parts such that their sum is $-3$ and their product is $2$. The numbers are $-2$ and $-1$:

$y^2-2y-y+2=0$

Factor out common terms from the pairs:

$y(y-2)-1(y-2)=0$

$(y-2)(y-1)=0$

Setting each factor to zero gives:

$y-2=0\Rightarrow y=2$

$y-1=0\Rightarrow y=1$

So, the roots for $y$ are $y=1$ and $y=2$.

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

Now, let us compare every calculated value of $x$ with every value of $y$:

• For $x=3$ and $y=1$: $3>1$ ($x>y$)

• For $x=3$ and $y=2$: $3>2$ ($x>y$)

• For $x=8$ and $y=1$: $8>1$ ($x>y$)

• For $x=8$ and $y=2$: $8>2$ ($x>y$)

Since every value of $x$ is strictly greater than every value of $y$, the final relationship is $x>y$.