Directions: Solve both quadratic equations and compare the possible values of x and y.

I. $x^2-3x+2=0$

II. $y^2-9y+20=0$

The correct answer is x < y.

To determine the relationship between $x$ and $y$, we solve both quadratic equations step-by-step and compare all possible values of $x$ with all possible values of $y$.

Step 1: Solve Equation I for $x$

The first equation is:

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

To factorize this quadratic equation, we find two numbers that multiply to $2$ and add up to $-3$. These numbers are $-1$ and $-2$.

Rewriting and factorizing the equation:

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

Setting each factor equal to zero:

$x-1=0\Rightarrow x=1$

$x-2=0\Rightarrow x=2$

Thus, the roots of the first equation are $x=1$ and $x=2$.

Step 2: Solve Equation II for $y$

The second equation is:

$y^2-9y+20=0$

To factorize this quadratic equation, we find two numbers that multiply to $20$ and add up to $-9$. These numbers are $-4$ and $-5$.

Rewriting and factorizing the equation:

$(y-4)(y-5)=0$

Setting each factor equal to zero:

$y-4=0\Rightarrow y=4$

$y-5=0\Rightarrow y=5$

Thus, the roots of the second equation are $y=4$ and $y=5$.

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

Now, we compare each value of $x$ with each value of $y$:

• For $x=1$ and $y=4$: $1<4$ ($x<\mathrm{y}$)

• For $x=1$ and $y=5$: $1<5$ ($x<\mathrm{y}$)

• For $x=2$ and $y=4$: $2<4$ ($x<\mathrm{y}$)

• For $x=2$ and $y=5$: $2<5$ ($x<\mathrm{y}$)

Since every possible value of $x$ is strictly less than every possible value of $y$, we conclude that $x<\mathrm{y}$.

Therefore, the correct relationship is x < y.