Given the two quadratic equations:

I. $(2x^2 + 7x - 15 = 0)$

II. $(3y^2 + 5y - 12 = 0)$

Determine the correct relationship between $x$ and $y$ from the following options:

Question

Given the two quadratic equations:

I. $(2x^2 + 7x - 15 = 0)$

II. $(3y^2 + 5y - 12 = 0)$

Determine the correct relationship between $x$ and $y$ from the following options:

Accepted Answer

**Solution:**

1. Solve the first equation:

$$2x^2 + 7x - 15 = 0$$

Factor by grouping:

$$2x^2 + 10x - 3x - 15 = 0$$
$$2x(x + 5) - 3(x + 5) = 0$$
$$(x + 5)(2x - 3) = 0$$

So, $x = -5$ or $x = \frac{3}{2}$.

2. Solve the second equation:

$$3y^2 + 5y - 12 = 0$$

Factor by grouping:

$$3y^2 + 9y - 4y - 12 = 0$$
$$3y(y + 3) - 4(y + 3) = 0$$
$$(y + 3)(3y - 4) = 0$$

So, $y = -3$ or $y = \frac{4}{3}$.

3. Compare possible values of $x$ and $y$:

- $x$ can be $-5$ or $1.5$
- $y$ can be $-3$ or $1.333...$

Since $x$ and $y$ can take multiple values, and no consistent inequality holds for all pairs, no definite relation between $x$ and $y$ can be established.

**Answer:** (e) $x = y$ or no definite relation can be established

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