Question

Given the equation \(x^2 - 5x = 6 - y^2 + 3y = 9y + 9\).
Quantity A: \(x\)
Quantity B: \(y\)

• A. if Quantity A is greater;
• B. if Quantity B is greater;
• C. if the two quantities are equal;
• D. if the relationship cannot be determined from the information given.
• E.

Hint:

In quantitative comparison questions, if you find that a variable can take on multiple values, test different possibilities. If you can show that in one scenario Quantity A is greater, and in another scenario Quantity B is greater, then the answer is always (D).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem provides a compound equality involving two variables, \(x\) and \(y\). We need to determine if there is a consistent relationship between \(x\) and \(y\). A compound equality \(A = B = C\) can be split into two separate equations, such as \(A = C\) and \(B = C\).
Step 2: Key Formula or Approach:
We will first solve for the possible values of \(y\) by setting the second and third expressions equal to each other. This will result in a quadratic equation, which can be solved using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Then we will analyze the resulting values for \(x\).
Step 3: Detailed Explanation:
Let's first find the value(s) of \(y\). We use the equality \(6 - y^2 + 3y = 9y + 9\):
\[ 6 - y^2 + 3y = 9y + 9 \]
Rearrange the terms to form a standard quadratic equation (\(ay^2 + by + c = 0\)):
\[ 0 = y^2 + 9y - 3y + 9 - 6 \]
\[ y^2 + 6y + 3 = 0 \]
Using the quadratic formula to solve for \(y\):
\[ y = \frac{-6 \pm \sqrt{6^2 - 4(1)(3)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 12}}{2} = \frac{-6 \pm \sqrt{24}}{2} = \frac{-6 \pm 2\sqrt{6}}{2} \]
\[ y = -3 \pm \sqrt{6} \]
This gives two possible values for \(y\): \(y_1 = -3 + \sqrt{6}\) and \(y_2 = -3 - \sqrt{6}\).
Now let's consider the equation for \(x\): \(x^2 - 5x = 9y + 9\).
For each value of \(y\), we will get a quadratic equation for \(x\), which will yield two values for \(x\). Let's test if we can find contradictory relationships.
Case 1: Let's use \(y_1 = -3 + \sqrt{6}\).
We know that \(\sqrt{4} < \sqrt{6} < \sqrt{9}\), so \(2 < \sqrt{6} < 3\). Let's approximate \(\sqrt{6} \approx 2.45\).
So, \(y_1 \approx -3 + 2.45 = -0.55\).
The equation for \(x\) becomes \(x^2 - 5x = 9(-3 + \sqrt{6}) + 9 = -27 + 9\sqrt{6} + 9 = 9\sqrt{6} - 18\).
So, \(x^2 - 5x - (9\sqrt{6} - 18) = 0\).
The solutions for \(x\) are given by: \[ x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-(9\sqrt{6}-18))}}{2} = \frac{5 \pm \sqrt{25 + 36\sqrt{6} - 72}}{2} = \frac{5 \pm \sqrt{36\sqrt{6} - 47}}{2} \]
Let's approximate the term under the square root: \(36\sqrt{6} - 47 \approx 36(2.45) - 47 = 88.2 - 47 = 41.2\).
So, \(x \approx \frac{5 \pm \sqrt{41.2}}{2} \approx \frac{5 \pm 6.42}{2}\).
One possible value for \(x\) is \(x_1 \approx \frac{5 + 6.42}{2} = 5.71\). In this instance, \(x_1 \approx 5.71 > y_1 \approx -0.55\).
Another possible value for \(x\) is \(x_2 \approx \frac{5 - 6.42}{2} = -0.71\). In this instance, \(x_2 \approx -0.71 < y_1 \approx -0.55\).
Step 4: Final Answer:
We have found one pair of values \((x_1, y_1)\) where \(x > y\) and another pair \((x_2, y_1)\) where \(x < y\). Since the relationship between \(x\) and \(y\) is not constant and depends on which solution we consider, the relationship cannot be determined from the information given.