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Question

If $x = \frac{\sqrt{9}+1}{\sqrt{9}-1}$ and $y = \frac{\sqrt{16}+1}{\sqrt{16}-1}$, then the value of $(x^2 - y^2)$ is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{11}{9}$

Value of $(x^2 - y^2)$ Calculation

This solution finds the value of the expression $(x^2 - y^2)$ using the given definitions for $x$ and $y$. The process involves simplifying the terms, squaring them, and then finding their difference.

$x$ Value Simplification

First, simplify the expression for $x$: $x = \frac{\sqrt{9}+1}{\sqrt{9}-1}$ Since $\sqrt{9} = 3$, we substitute this value: $x = \frac{3+1}{3-1} = \frac{4}{2}$ Thus, $x = 2$.

$y$ Value Simplification

Next, simplify the expression for $y$: $y = \frac{\sqrt{16}+1}{\sqrt{16}-1}$ Since $\sqrt{16} = 4$, we substitute this value: $y = \frac{4+1}{4-1} = \frac{5}{3}$ Thus, $y = \frac{5}{3}$.

Calculate Squares $x^2$ and $y^2$

Now, calculate the squares of $x$ and $y$: $x^2 = 2^2 = 4$ $y^2 = (\frac{5}{3})^2 = \frac{25}{9}$

Difference $x^2 - y^2$ Calculation

Finally, calculate the difference $(x^2 - y^2)$: $x^2 - y^2 = 4 - \frac{25}{9}$ To subtract, find a common denominator, which is 9: $x^2 - y^2 = \frac{4 \times 9}{9} - \frac{25}{9} = \frac{36}{9} - \frac{25}{9}$ $x^2 - y^2 = \frac{36 - 25}{9}$ $x^2 - y^2 = \frac{11}{9}$

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