Solutions For System Of Equations X^2 + Y^2 = 6 And X^2 - Y = 6

In mathematics, solving a system of equations is a fundamental skill. This article delves into finding the solutions for a specific system of equations. We'll explore the methods, step-by-step solutions, and potential pitfalls in solving such problems. Our focus will be on a system involving both a quadratic equation and a linear-like equation. Understanding how to approach these problems is crucial for various mathematical applications and further studies in algebra and calculus.

Let's consider the following system of equations:

x^2 + y^2 = 6
x^2 - y = 6

Our goal is to determine the solution(s) for this system. This means finding the pair(s) of (x, y) values that satisfy both equations simultaneously. The given options are:

A. no solution B. $(\sqrt{6}, 0)$ C. $(\sqrt{5}, 1)$ D. $(\sqrt{5}, -1)$ E. infinitely many

We will use algebraic techniques to find the correct solution(s) and eliminate the incorrect ones. This process will involve substitution, simplification, and potentially dealing with quadratic equations. Accurate manipulation of these equations is key to arriving at the right answer.

Method 1: Substitution

To solve this system of equations, the most effective method is often substitution. Substitution involves solving one equation for one variable and then substituting that expression into the other equation. In our case, the second equation, $x^2 - y = 6$, is easily solved for $x^2$. Let's isolate $x^2$ in the second equation:

$x^2 = y + 6$

Now, we can substitute this expression for $x^2$ into the first equation, $x^2 + y^2 = 6$:

$(y + 6) + y^2 = 6$

This substitution results in a quadratic equation in terms of $y$. Simplifying this equation will allow us to solve for the possible values of $y$. This is a crucial step in finding the solutions to the system, as it reduces the problem to a single variable equation. The resulting equation will help us determine the y-coordinates of the solutions, which we can then use to find the corresponding x-coordinates.

Solving the Quadratic Equation

After substitution, we have the equation:

$y^2 + y + 6 = 6$

Subtracting 6 from both sides, we get:

$y^2 + y = 0$

This quadratic equation can be easily factored:

$y(y + 1) = 0$

This gives us two possible solutions for $y$:

$y = 0$ or $y = -1$

These values of $y$ are the potential y-coordinates of the solutions to our system of equations. We must now determine the corresponding $x$ values for each of these $y$ values. This involves substituting each $y$ value back into one of the original equations to solve for $x$. By finding the corresponding $x$ values, we can identify the pairs $(x, y)$ that satisfy both equations in the system.

Finding the Corresponding x Values

Now that we have the $y$ values, we can find the corresponding $x$ values. We'll substitute each $y$ value back into the equation $x^2 = y + 6$, which we derived earlier.

For $y = 0$:

$x^2 = 0 + 6$ $x^2 = 6$ $x = \pm\sqrt{6}$

This gives us two solutions: $(\sqrt{6}, 0)$ and $(-\sqrt{6}, 0)$.

For $y = -1$:

$x^2 = -1 + 6$ $x^2 = 5$ $x = \pm\sqrt{5}$

This gives us two more solutions: $(\sqrt{5}, -1)$ and $(-\sqrt{5}, -1)$.

So, we have found four potential solutions to the system of equations. It is essential to check each of these solutions in both original equations to ensure they are valid and not extraneous solutions introduced during the algebraic manipulation.

Verification of Solutions

It's crucial to verify our solutions by substituting them back into the original equations. This step ensures that the solutions we found are valid and haven't been introduced by any algebraic manipulations.

Let's check each of our potential solutions:

1. $(\sqrt{6}, 0)$:

   • $(\sqrt{6})^2 + 0^2 = 6 + 0 = 6$ (satisfies the first equation)
   • $(\sqrt{6})^2 - 0 = 6 - 0 = 6$ (satisfies the second equation)

2. $(-\sqrt{6}, 0)$:

   • $(-\sqrt{6})^2 + 0^2 = 6 + 0 = 6$ (satisfies the first equation)
   • $(-\sqrt{6})^2 - 0 = 6 - 0 = 6$ (satisfies the second equation)

3. $(\sqrt{5}, -1)$:

   • $(\sqrt{5})^2 + (-1)^2 = 5 + 1 = 6$ (satisfies the first equation)
   • $(\sqrt{5})^2 - (-1) = 5 + 1 = 6$ (satisfies the second equation)

4. $(-\sqrt{5}, -1)$:

   • $(-\sqrt{5})^2 + (-1)^2 = 5 + 1 = 6$ (satisfies the first equation)
   • $(-\sqrt{5})^2 - (-1) = 5 + 1 = 6$ (satisfies the second equation)

All four solutions satisfy both equations. Therefore, these are the valid solutions to the system.

Now, let's analyze the given options based on our solutions:

A. no solution: This is incorrect as we found four solutions. B. $(\sqrt{6}, 0)$: This is one of the solutions. C. $(\sqrt{5}, 1)$: This is incorrect. While $x = \sqrt{5}$ is part of a solution, the corresponding $y$ value is $-1$, not $1$. D. $(\sqrt{5}, -1)$: This is one of the solutions. E. infinitely many: This is incorrect as we found a finite number of solutions.

Therefore, the correct solutions are B and D.

The solutions to the system of equations are $(\sqrt{6}, 0)$ and $(\sqrt{5}, -1)$. These points are where the graphs of the two equations intersect. The first equation, $x^2 + y^2 = 6$, represents a circle centered at the origin with a radius of $\sqrt{6}$. The second equation, $x^2 - y = 6$, represents a parabola opening downwards. The points of intersection of these two curves correspond to the solutions of the system. Understanding the graphical representation of the equations can provide a visual confirmation of the solutions we found algebraically.

In conclusion, solving a system of equations often involves multiple steps, including substitution, simplification, and verification. For the given system:

x^2 + y^2 = 6
x^2 - y = 6

We found four solutions: $(\sqrt{6}, 0)$, $(-\sqrt{6}, 0)$, $(\sqrt{5}, -1)$, and $(-\sqrt{5}, -1)$. The correct options from the list provided were B and D. This exercise highlights the importance of careful algebraic manipulation and the need to verify solutions to ensure accuracy. Mastering these techniques is essential for success in algebra and related fields.