In each of these questions, two equations (I) and (II) are given. Solve the equations and mark the correct option.

I. 6x² + x- 15 =0
II. 4y² - 24y + 35 = 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 first equation is:

$6x^2+x-15=0$

We can solve this by factorizing the quadratic expression. We look for two numbers whose product is $6\times (-15)=-90$ and whose sum is $1$. These numbers are $10$ and $-9$.

Split the middle term:

$6x^2+10x-9x-15=0$

Factor by grouping:

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

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

Equating each factor to zero gives:

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

$3x+5=0\Rightarrow x=-\frac{5}{3}\approx -1.67$

Thus, the roots for $x$ are $x=1.5$ and $x=-1.67$.

Step 2: Solve Equation II for $y$

The second equation is:

$4y^2-24y+35=0$

We find two numbers whose product is $4\times 35=140$ and whose sum is $-24$. These numbers are $-14$ and $-10$.

Split the middle term:

$4y^2-14y-10y+35=0$

Factor by grouping:

$2y(2y-7)-5(2y-7)=0$

$(2y-5)(2y-7)=0$

Equating each factor to zero gives:

$2y-5=0\Rightarrow y=\frac{5}{2}=2.5$

$2y-7=0\Rightarrow y=\frac{7}{2}=3.5$

Thus, the roots for $y$ are $y=2.5$ and $y=3.5$.

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

Let us compare every possible pair of $x$ and $y$ values:

• For $x=1.5$: both $1.5<2.5$ and $1.5<3.5$ ($x<y$)

• For $x=-1.67$: both $-1.67<2.5$ and $-1.67<3.5$ ($x<y$)

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

Therefore, the relationship is $x<y$.