Directions: In the following question, two equations numbered I and II are given. Solve both equations and establish the correct relationship between x and y.

$\begin{array}{ll}\text{I.}& 2x^2-13x+15=0\\ \text{II.}& y^2-12y+36=0\end{array}$

Correct Answer: If x < y

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

Solving Equation I:

$2x^2-13x+15=0$

We factor the quadratic expression by splitting the middle term $-13x$ into two terms whose sum is $-13$ and product is $2\times 15=30$. The numbers are $-10$ and $-3$.

$2x^2-10x-3x+15=0$

Group the terms to factor out common factors:

$2x(x-5)-3(x-5)=0$

$(2x-3)(x-5)=0$

Set each factor equal to zero:

$2x-3=0\Rightarrow x=\frac{3}{2}=1.5$

$x-5=0\Rightarrow x=5$

Thus, the roots for x are $x=1.5$ and $x=5$.

Solving Equation II:

$y^2-12y+36=0$

Recognize that this is a perfect square trinomial:

$(y-6{)}^2=0$

$y-6=0\Rightarrow y=6$

Thus, the root for y is $y=6$.

Comparison of x and y values:

Comparing $x=1.5$ with $y=6$: $1.5<6$ ($x<y$).
Comparing $x=5$ with $y=6$: $5<6$ ($x<y$).

Since both values of x are strictly less than y, the correct relationship is If x < y.