If $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy} ext{ and } \frac{4}{x} + \frac{9}{y} = \frac{21}{xy}$, where x ≠ 0, y ≠ 0, then what is the value of x + y ?

Question

If $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy} 	ext{ and } \frac{4}{x} + \frac{9}{y} = \frac{21}{xy}$, where x ≠ 0, y ≠ 0, then what is the value of x + y ?

Accepted Answer

To solve the given problem and verify why Option 3 is correct, we need to analyze the system of equations:

1. $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}$
2. $\frac{4}{x} + \frac{9}{y} = \frac{21}{xy}$

We are tasked with finding the value of $x + y$.

**Step-by-step Explanation:**

Let's start by simplifying and solving these equations.

**Equation 1:**
$$
\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}
$$
Multiply through by $xy$ to eliminate the fractions:
$$
2y + 3x = 9
$$

**Equation 2:**
$$
\frac{4}{x} + \frac{9}{y} = \frac{21}{xy}
$$
Multiply through by $xy$ to eliminate the fractions:
$$
4y + 9x = 21
$$

Now, we have a system of linear equations:
1. $2y + 3x = 9$
2. $4y + 9x = 21$

To solve this system, we can use the method of elimination. Multiply the first equation by 2 to align the coefficients of $y$:

$$
4y + 6x = 18
$$

Now subtract the first equation from the second:
$$
(4y + 9x) - (4y + 6x) = 21 - 18
$$
$$
3x = 3
$$
$$
x = 1
$$

Substitute $x = 1$ back into the first equation:
$$
2y + 3(1) = 9
$$
$$
2y + 3 = 9
$$
$$
2y = 6
$$
$$
y = 3
$$

Thus, we have $x = 1$ and $y = 3$. Therefore, the value of $x + y$ is:
$$
x + y = 1 + 3 = 4
$$

**Conclusion:**

The correct value of $x + y$ is 4, which corresponds to Option 3.

**Explanation of Incorrect Options:**

- **Option 1 (2):** This would imply $x + y = 2$, which is not possible given the solutions $x = 1$ and $y = 3$.
- **Option 2 (3):** This would imply $x + y = 3$, which is incorrect as shown by our calculations.
- **Option 4 (8):** This would imply $x + y = 8$, which is also incorrect based on the derived values of $x$ and $y$.

Therefore, the only correct answer is Option 3, where $x + y = 4$.