Animated Solution for Mathematics - Quadratic Equations: If $x<0,y<0,x+y+\frac{x}{y}=\frac{1}{2}$ and $(x+y)\frac{x}{y}=-\frac{1}{2}$, then $x=\dots$ and $y=\dots$

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, might look like a messy tangle of variables.

You see $x+y+\frac{x}{y}=\frac{1}{2}$ and $(x+y)\cdot \frac{x}{y}=-\frac{1}{2}$. Your first instinct might be to isolate $x$ or $y$ and start substituting. But stop. Take a breath.

In JEE Advanced, the most elegant solution is rarely the brute-force one. Look closer at the structure. Do you see the repetition?

We have the sum $(x+y)$ and the ratio $\frac{x}{y}$ appearing in both equations. This is a classic signal to use the technique of 'masking' or substitution.

Let us define two new variables to simplify our lives. Let $a=x+y$ and $b=\frac{x}{y}$.

Suddenly, the problem transforms into a beautiful, symmetric system:

1. $a+b=\frac{1}{2}$

2. $a\cdot b=-\frac{1}{2}$

Now, we are no longer looking at $x$ and $y$. We are looking at the sum and product of two numbers, $a$ and $b$. This is the golden key!

According to Vieta's formulas, if we know the sum and product of two roots, those roots must be the solutions to a quadratic equation of the form $t^2-(\text{sum})t+(\text{product})=0$.

Substituting our values, we get the quadratic equation:

To make this easier to handle, let us clear the fraction by multiplying the entire equation by $2$:

Now, we factorize. We are looking for two numbers that multiply to $-2$ and add to $-1$. Those numbers are $-2$ and $1$.

So, we rewrite the middle term:

This gives us two potential values for our variables: $t=1$ and $t=-\frac{1}{2}$. This means our pair $(a,b)$ could be $(1,-\frac{1}{2})$ or $(-\frac{1}{2},1)$.

But here is where the JEE examiner tests your attention to detail. We were given the constraint $x<0$ and $y<0$.

If both $x$ and $y$ are negative, their sum $a=x+y$ must also be negative. Therefore, $a$ cannot be $1$.

We must reject the pair where $a=1$. The only valid pair is $a=-\frac{1}{2}$ and $b=1$.

With $a=-\frac{1}{2}$ and $b=1$, we are in the home stretch. Since $b=\frac{x}{y}=1$, we know that $x=y$.

And since $a=x+y=-\frac{1}{2}$, we can substitute $y$ with $x$ to get:

Since $x=y$, it follows that $y=-\frac{1}{4}$. We have arrived at our destination.

The beauty of this problem lies not in the calculation, but in the recognition of the structure and the disciplined application of constraints. Keep this mindset—look for the pattern, build the bridge, and respect the constraints—and you will conquer any problem the exam throws at you. The final result is $x=-\frac{1}{4},y=-\frac{1}{4}$.