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

I. $x^2-8x+15=0$

II. $y^2-3y+2=0$

Correct Answer: If x > y

Step-by-step Explanation:

Step 1: Solve Equation I for x

The first quadratic equation is:

$x^2-8x+15=0$

To factorize this quadratic equation, we need two numbers that multiply to 15 and add up to -8. These numbers are -3 and -5.

$x^2-3x-5x+15=0$

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

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

Setting each factor equal to zero:

$x-3=0$ ⇒ $x=3$

$x-5=0$ ⇒ $x=5$

So, the values of x are 3 and 5.

Step 2: Solve Equation II for y

The second quadratic equation is:

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

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

$y^2-1y-2y+2=0$

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

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

Setting each factor equal to zero:

$y-1=0$ ⇒ $y=1$

$\mathrm{y<mo>-</mo><mn>2</mn><mo>=</mo><mn>0</mn>}$ ⇒ $y=2$

So, the values of y are 1 and 2.

Step 3: Compare the values of x and y

Now, let us compare every calculated value of x with every calculated value of y:

When $x=3$:

3 > 1 and 3 > 2 (x is greater than both values of y)

When $x=5$:

5 > 1 and 5 > 2 (x is greater than both values of y)

Since every value of x is strictly greater than every value of y, the correct relationship between x and y is If x > y.