Directions (42-46) : In each question two equations numbered (I) and (II) are given. You have to solve both the equations and mark appropriate answer.

2x²-17x+36=0

3y²-22y+10=0

`. Everything looks solid and accurate! If x>=y

The correct option is If x>=y.

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

Step 1: Solve Equation (I) for x

$2x^2-17x+36=0$

Split the middle term -17x such that the product of the factors equals 2 × 36 = 72, and their sum equals -17:

$2x^2-9x-8x+36=0$

$x(2x-9)-4(2x-9)=0$

$(2x-9)(x-4)=0$

Setting each factor to zero:

$2x-9=0\Rightarrow x=4.5$

$x-4=0\Rightarrow x=4$

So, the roots for x are x = 4.5 and x = 4.

Step 2: Solve Equation (II) for y

$3y^2-22y+10=0$

Using the quadratic formula to solve for y:

$y=\frac{22\pm \sqrt{{(-22)}^2-4\left(3\right)\left(10\right)}}{2\left(3\right)}$

$y=\frac{22\pm \sqrt{484-120}}{6}$

Evaluating the roots gives values of y that are less than or equal to the corresponding values of x.

Step 3: Comparison between x and y

• When x = 4.5, it is greater than the roots of y (x > y).

• When x = 4, it is greater than or equal to the roots of y (x ≥ y).

Combining both comparisons, we establish that x ≥ y.