Solving $2y + X = 1$ And $x^2 + Y^2 + 3xy + Y = 0$ A Step-by-Step Guide

Hey guys! Let's dive into solving a cool system of equations. We've got two equations here: $2y + x = 1$ and $x^2 + y^2 + 3xy + y = 0$. It looks a bit intimidating at first, but don't worry, we'll break it down step by step.

Understanding the Equations

Before we jump into solving, let's take a closer look at what we're dealing with. The first equation, $2y + x = 1$, is a linear equation. This means it represents a straight line when graphed. Linear equations are generally easier to work with because they have a predictable form. We can easily isolate one variable in terms of the other. For instance, we can rewrite this equation as $x = 1 - 2y$, which will be super helpful later on when we use substitution.

The second equation, $x^2 + y^2 + 3xy + y = 0$, is a bit more complex. It's a quadratic equation, meaning it involves terms with variables raised to the power of 2. The presence of both $x^2$ and $y^2$ terms, as well as the $3xy$ term, suggests that this equation might represent a conic section (like an ellipse, hyperbola, or parabola) or a more complicated curve. The $3xy$ term is particularly important because it indicates that the curve is rotated relative to the standard conic sections. This added complexity means we need a clever approach to solve the system.

When we have a system of equations like this, where one equation is linear and the other is quadratic, the substitution method is often the most effective strategy. The basic idea behind substitution is to solve the linear equation for one variable and then substitute that expression into the quadratic equation. This will give us a new equation in just one variable, which we can then solve using techniques like factoring or the quadratic formula. Once we've found the values of one variable, we can plug them back into either of the original equations to find the corresponding values of the other variable. This process might sound a bit abstract right now, but it'll become clearer as we work through the problem step by step.

Step-by-Step Solution

Alright, let's get our hands dirty and start solving this system. As we discussed earlier, the first step is to solve the linear equation, $2y + x = 1$, for one of the variables. It looks simpler to solve for $x$, so let's do that. Subtracting $2y$ from both sides, we get:

$x = 1 - 2y$

Great! Now we have an expression for $x$ in terms of $y$. The next step is to substitute this expression into the second equation, $x^2 + y^2 + 3xy + y = 0$. This might seem a bit messy, but hang in there. We're replacing every instance of $x$ in the second equation with $(1 - 2y)$. So, we get:

(1 - 2y)^2 + y^2 + 3(1 - 2y)y + y = 0

Okay, we've successfully substituted! Now we need to simplify this equation. Let's start by expanding the terms. First, $(1 - 2y)^2$ expands to $(1 - 4y + 4y^2)$. And $3(1 - 2y)y$ expands to $3y - 6y^2$. Substituting these back into the equation, we get:

1 - 4y + 4y^2 + y^2 + 3y - 6y^2 + y = 0

Now we can combine like terms. Let's group the $y^2$ terms, the $y$ terms, and the constant terms together:

(4y^2 + y^2 - 6y^2) + (-4y + 3y + y) + 1 = 0

Simplifying this, we get:

-y^2 + 0y + 1 = 0

Which simplifies further to:

-y^2 + 1 = 0

Or, equivalently:

y^2 = 1

This is a much simpler equation to solve! Taking the square root of both sides, we get two possible values for $y$:

$y = 1$ or $y = -1$

We're halfway there! We've found the possible values for $y$. Now we need to find the corresponding values for $x$. This is where we go back to our expression for $x$ in terms of $y$, which is $x = 1 - 2y$.

Let's start with $y = 1$. Plugging this into the equation, we get:

$x = 1 - 2(1) = 1 - 2 = -1$

So, one solution is $x = -1$ and $y = 1$. This gives us the ordered pair $(-1, 1)$.

Now let's consider the other value, $y = -1$. Plugging this into the equation $x = 1 - 2y$, we get:

$x = 1 - 2(-1) = 1 + 2 = 3$

So, the other solution is $x = 3$ and $y = -1$. This gives us the ordered pair $(3, -1)$.

Therefore, the solutions to the system of equations are $(-1, 1)$ and $(3, -1)$. We've successfully solved the system! Pat yourselves on the back, guys. That was a bit of a journey, but we made it through.

Verification and Pitfalls

Now, it's always a good idea to verify our solutions. This means plugging the $x$ and $y$ values we found back into the original equations to make sure they satisfy both equations. This helps us catch any mistakes we might have made along the way.

Let's start with the solution $(-1, 1)$. Plugging these values into the first equation, $2y + x = 1$, we get:

$2(1) + (-1) = 2 - 1 = 1$

This checks out! Now let's plug these values into the second equation, $x^2 + y^2 + 3xy + y = 0$:

$(-1)^2 + (1)^2 + 3(-1)(1) + 1 = 1 + 1 - 3 + 1 = 0$

This also checks out! So, the solution $(-1, 1)$ is definitely correct.

Now let's verify the second solution, $(3, -1)$. Plugging these values into the first equation, $2y + x = 1$, we get:

$2(-1) + 3 = -2 + 3 = 1$

This is correct as well. Plugging these values into the second equation, $x^2 + y^2 + 3xy + y = 0$, we get:

$(3)^2 + (-1)^2 + 3(3)(-1) + (-1) = 9 + 1 - 9 - 1 = 0$

This solution also checks out! So, we've confirmed that both $(-1, 1)$ and $(3, -1)$ are valid solutions to the system of equations.

One common pitfall when solving systems of equations is making algebraic errors during the substitution and simplification steps. It's really easy to drop a negative sign or make a mistake when expanding terms. That's why it's so important to be careful and methodical in our calculations. Another common mistake is forgetting to solve for both variables. Remember, when we use substitution, we initially find the value(s) of one variable, but we still need to plug those values back in to find the corresponding values of the other variable. Failing to do this will leave us with only half the solution.

Alternative Approaches

While the substitution method worked great for this system, it's worth noting that there might be other approaches we could use. For example, if the equations had a different structure, we might consider using elimination. Elimination involves manipulating the equations so that when we add or subtract them, one of the variables cancels out. This can be a powerful technique, especially when dealing with systems of linear equations.

Another approach, although less practical for this specific problem, is to use graphical methods. We could graph both equations on the same coordinate plane and look for the points where the graphs intersect. These intersection points represent the solutions to the system. While this method can be visually helpful, it's not always accurate, especially when the solutions are not integers.

In some cases, we might also consider using numerical methods, especially if the equations are very complex and don't have a simple algebraic solution. Numerical methods involve using iterative algorithms to approximate the solutions to a desired degree of accuracy. These methods are often implemented using computers or calculators.

However, for this particular system, the substitution method was the most straightforward and efficient approach. It allowed us to systematically reduce the problem to a single quadratic equation, which we could then solve easily.

Conclusion

So, there you have it! We've successfully solved the system of equations $2y + x = 1$ and $x^2 + y^2 + 3xy + y = 0$. We found two solutions: $(-1, 1)$ and $(3, -1)$. We used the substitution method, which involved solving the linear equation for one variable and then substituting that expression into the quadratic equation. We also verified our solutions to make sure they were correct. Remember, solving systems of equations is a fundamental skill in mathematics, and it's something you'll encounter in many different contexts. Keep practicing, and you'll become a pro in no time! Great job, everyone!