Given the two equations:

I. $(2x^2 + 26x + 84 = 0)$

II. $(y^2 + 15y + 56 = 0)$

Determine the correct relationship between $x$ and $y$:

Question

Given the two equations:

I. $(2x^2 + 26x + 84 = 0)$

II. $(y^2 + 15y + 56 = 0)$

Determine the correct relationship between $x$ and $y$:

Accepted Answer

**Solution:**

1. First, simplify and solve the equation for $x$:

$$ 2x^2 + 26x + 84 = 0 $$

Divide the entire equation by 2:

$$ x^2 + 13x + 42 = 0 $$

2. Factorize the quadratic:

$$ x^2 + 6x + 7x + 42 = 0 $$

Group terms:

$$ x(x + 6) + 7(x + 6) = 0 $$

$$ (x + 6)(x + 7) = 0 $$

So, $ x = -6 $ or $ x = -7 $.

3. Now, solve the equation for $y$:

$$ y^2 + 15y + 56 = 0 $$

Factorize:

$$ y^2 + 7y + 8y + 56 = 0 $$

Group terms:

$$ y(y + 7) + 8(y + 7) = 0 $$

$$ (y + 7)(y + 8) = 0 $$

So, $ y = -7 $ or $ y = -8 $.

4. Compare the roots:

- $x$ can be $-6$ or $-7$
- $y$ can be $-7$ or $-8$

5. Check all possible comparisons:

- $x = -6$ and $y = -7$ → $x > y$
- $x = -6$ and $y = -8$ → $x > y$
- $x = -7$ and $y = -7$ → $x = y$
- $x = -7$ and $y = -8$ → $x > y$

In all cases, $x \geq y$.

**Answer:** (c) $x \geq y$

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