Directions: In each of the following questions, two equations numbered (I) and (II) are given. Solve both equations and select the appropriate option.

I. $6x^2-17x+12=0$
II. $6y^2-7y+2=0$

The correct option is If x > y.

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

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

The given equation is:

$6x^2-17x+12=0$

We split the middle term $-17x$ into two terms whose product is $6\times 12=72$ and whose sum is $-17$. These factors are $-9$ and $-8$.

$6x^2-9x-8x+12=0$

Factorize by grouping:

$3x(2x-3)-4(2x-3)=0$

$(3x-4)(2x-3)=0$

Setting each factor equal to zero gives:

$3x-4=0\Rightarrow x=\frac{4}{3}\approx 1.33$

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

Thus, the roots for $x$ are $\frac{4}{3}$ and $\frac{3}{2}$ (or approximately $1.33$ and $1.5$).

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

The given equation is:

$6y^2-7y+2=0$

We split the middle term $-7y$ into two terms whose product is $6\times 2=12$ and whose sum is $-7$. These factors are $-3$ and $-4$.

$6y^2-3y-4y+2=0$

Factorize by grouping:

$3y(2y-1)-2(2y-1)=0$

$(3y-2)(2y-1)=0$

Setting each factor equal to zero gives:

$3y-2=0\Rightarrow y=\frac{2}{3}\approx 0.67$

$2y-1=0\Rightarrow y=\frac{1}{2}=0.5$

Thus, the roots for $y$ are $\frac{2}{3}$ and $\frac{1}{2}$ (or approximately $0.67$ and $0.5$).

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

Comparing each calculated value of $x$ with each value of $y$:

• For $x=1.33$ and $y=0.67$: $1.33>0.67$ ($x>y$)

• For $x=1.33$ and $y=0.5$: $1.33>0.5$ ($x>y$)

• For $x=1.5$ and $y=0.67$: $1.5>0.67$ ($x>y$)

• For $x=1.5$ and $y=0.5$: $1.5>0.5$ ($x>y$)

In all cases, every value of $x$ is strictly greater than every value of $y$.

Hence, the correct relationship is If x > y.