CAT 2023 Slot 1 QA Question 3

First, the equation given is:

Hmm, okay. So, We find $x - 2y$. how I can manipulate this equation to get that expression.

Note that both sides of the equation have terms involving $x$ and $y$. Maybe expanding the left side will help. Expanding $(x - 2y - 1)^2$: $(x - 2y - 1)^2 = (x - 2y)^2 - 2(x - 2y)(1) + 1^2$ = x^2 - 4xy + 4y^2 - 2x + 4y + 1$

So, substituting back into the original equation:

Let me combine like terms on the left side: $x^2 + x^2 - 4xy + 4y^2 - 2x + 4y + 1$ = 2x^2 - 4xy + 4y^2 - 2x + 4y + 1$

Now, the right side is: $- 4y(x + y) = -4xy - 4y^2$

So, putting it all together:

Let me bring all terms to the left side:

Simplify by combining like terms:

So, the equation becomes:

Hmm, this looks a bit complicated. Maybe I can divide the entire equation by 2 to simplify:

Still, it's not very straightforward. Maybe completing the square would help here. Let me try that for both $x$ and $y$.

Starting with the $x$ terms: $x^2 - x$ To complete the square, take half the coefficient of $x , which is - \frac{1}{2} , square it to get \frac{1}{4}$. So: $x^2 - x + \frac{1}{4} = \left(x - \frac{1}{2}\right)^2$$

Similarly, for the $y$ terms: $4y^2 + 2y$$Factor out 4 first:$$4\left(y^2 + \frac{1}{2}y\right)$ Now, complete the square inside the parentheses. Half of $\frac{1}{2}$ is $\frac{1}{4}$ , squared is $\frac{1}{16}$$. So:$4\left(y^2 + $\frac{1}{2}$y + $\frac{1}{16}$ - $\frac{1}{16}$\right)$= 4\left(\left(y + \frac{1}{4}\right)^2 - \frac{1}{16}\right)$ = 4\left(y + $\frac{1}{4}$\right)^2 - $\frac{1}{4}$$

Now, substitute these back into the equation:

Simplify the constants: $- \frac{1}{4} + \frac{1}{2} = \frac{1}{4}$ So, the equation becomes:

Wait a minute, the sum of squares plus a positive constant equals zero? That can't happen unless each square is zero and the constant is zero. But here, the constant is $\frac{1}{4}$ , which is positive. That suggests there's no real solution, but the problem states that $x$ and $y$ are real numbers. Hmm, maybe I made a mistake in my calculations.

Starting from the expansion:

Original equation:

Expanding $(x - 2y - 1)^2$: $x^2 - 4xy + 4y^2 - 2x + 4y + 1$$

Adding to $x^2$: $x^2 + x^2 - 4xy + 4y^2 - 2x + 4y + 1$ = 2x^2 - 4xy + 4y^2 - 2x + 4y + 1$

Right side: $- 4y(x + y) = -4xy - 4y^2$

Bringing all terms to the left: $2x^2 - 4xy + 4y^2 - 2x + 4y + 1 + 4xy + 4y^2 = 0$ $2x^2 + 8y^2 - 2x + 4y + 1 = 0$

Dividing by 2:

Completing the square for $x$: $x^2 - x = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4}$$

For $y$: \[ 4y^2 +