Solving Systems Of Equations: A Step-by-Step Guide

Hey guys! Today, we're diving into the world of solving systems of equations. Don't worry, it's not as scary as it sounds! We'll break down a specific example and walk through the steps, making sure you understand everything. Ready to get started? Let's go! This is a common problem in mathematics, especially in algebra. Understanding how to solve systems of equations is a fundamental skill, and it's super useful in various fields, from science and engineering to economics and computer science. So, let's learn how to tackle this type of problem. We will use two equations to solve the problem and also to teach you how to solve the problem systematically. In the end, you'll be able to solve similar systems of equations with confidence. The approach we'll be using is a combination of substitution and simplification. We'll strategically manipulate the equations to isolate variables and find the solutions. There are different methods to solve it, but in the end, it is important that we arrive at the correct answer. The more you practice, the better you get at recognizing the best approach for different types of systems. Now, let's jump into the details and work through the example step by step.

Understanding the Problem: The Equations at Hand

Alright, let's take a look at the system of equations we're going to solve. Here it is:

$\left\{\begin{array}{l} x^2+y^2=30 \\ x-y^2=-10 \end{array}\right.$

So, what does this actually mean? We have two equations, and we need to find the values of x and y that satisfy both equations simultaneously. Think of it like this: each equation represents a curve (in this case, a circle and a parabola), and the solution to the system is the point(s) where these curves intersect. Our goal is to find those intersection points. The first equation, $x^2 + y^2 = 30$, represents a circle centered at the origin (0, 0) with a radius of the square root of 30. The second equation, $x - y^2 = -10$, represents a parabola opening to the right. Therefore, the solution to the system will be the point(s) that lie on both the circle and the parabola. Before we start, let's take a moment to understand the structure of the equations. The first equation involves both $x^2$ and $y^2$, while the second equation includes x and $y^2$. This difference will be key in choosing our solution method. Knowing how the equations are structured will allow us to simplify the process and find the right way to solve the problem. Now, let's get down to business.

The Substitution Method: Our Strategy

The substitution method is a fantastic tool for solving systems of equations. The basic idea is to solve one equation for one variable and then substitute that expression into the other equation. This reduces the problem to a single equation with a single variable, which we can then solve. In our case, the second equation, $x - y^2 = -10$, is a perfect candidate for isolating a variable. It looks simpler than the first equation. We can easily solve it for x. This simplifies the problem because we will have an explicit expression for one variable in terms of the other. The next step will be to substitute the expression found in the other equation. It's like a chain reaction, which eventually helps us to find the final result of our problem. Let's see how this works in practice and how we can apply the substitution method to find the values of x and y. Remember that the goal is to simplify the system of equations step by step until we arrive at a solution.

Step 1: Isolating x

Let's isolate x in the second equation ($x - y^2 = -10$). To do this, we simply add $y^2$ to both sides of the equation: $x - y^2 + y^2 = -10 + y^2$, which simplifies to $x = y^2 - 10$. Great! Now we have an expression for x in terms of y. This is the first critical step in our solution. We've taken one of the equations and made it simpler, preparing it for the next phase. Now we know what x is equal to, so we can replace it in the first equation. This is the beauty of the substitution method. It allows us to transform a system of two equations into a simpler form that is easier to solve. The expression we found for x will now be integrated into the first equation.

Step 2: Substitution

Now, we'll substitute the expression we found for x ($x = y^2 - 10$) into the first equation ($x^2 + y^2 = 30$). This gives us: $(y^2 - 10)^2 + y^2 = 30$. See how we've replaced x with an equivalent expression? This is the heart of the substitution method. We now have a single equation with a single variable, y. It is important to remember that we substitute the value of x into the first equation. Now we can proceed to solve this equation for y. This will give us the values of y that satisfy both equations in our original system. Once we've found the values of y, we can go back and find the corresponding values of x. Let's move on and solve the resulting equation to find the value of y.

Step 3: Simplifying and Solving for y

Okay, let's simplify and solve the equation $(y^2 - 10)^2 + y^2 = 30$. First, expand the squared term: $(y^2 - 10)(y^2 - 10) + y^2 = 30$. This expands to $y^4 - 20y^2 + 100 + y^2 = 30$. Now, combine like terms: $y^4 - 19y^2 + 100 = 30$. Next, subtract 30 from both sides to set the equation to zero: $y^4 - 19y^2 + 70 = 0$. This is a quartic equation (an equation with a variable raised to the fourth power). However, it looks a lot like a quadratic equation. We can actually solve this by making a substitution. Let $z = y^2$. Then, our equation becomes: $z^2 - 19z + 70 = 0$. Now we can solve for z. This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. Let's try factoring. We're looking for two numbers that multiply to 70 and add up to -19. Those numbers are -14 and -5. So, we can factor the equation as $(z - 14)(z - 5) = 0$. This means that either $z - 14 = 0$ or $z - 5 = 0$. Solving for z, we get $z = 14$ or $z = 5$. But remember, we made the substitution $z = y^2$. Therefore, we have $y^2 = 14$ or $y^2 = 5$. To find the values of y, we take the square root of both sides. This gives us four possible solutions for y: $y = \sqrt{14}$, $y = -\sqrt{14}$, $y = \sqrt{5}$, and $y = -\sqrt{5}$.

Step 4: Finding the Values of x

Now that we have the values for y, let's find the corresponding values for x using the equation we derived earlier: $x = y^2 - 10$. Let's consider each value of y separately:

• If $y = \sqrt{14}$, then $x = (\sqrt{14})^2 - 10 = 14 - 10 = 4$. So, one solution is $(4, \sqrt{14})$.
• If $y = -\sqrt{14}$, then $x = (-\sqrt{14})^2 - 10 = 14 - 10 = 4$. So, another solution is $(4, -\sqrt{14})$.
• If $y = \sqrt{5}$, then $x = (\sqrt{5})^2 - 10 = 5 - 10 = -5$. So, another solution is $(-5, \sqrt{5})$.
• If $y = -\sqrt{5}$, then $x = (-\sqrt{5})^2 - 10 = 5 - 10 = -5$. So, another solution is $(-5, -\sqrt{5})$.

We found four solutions: $(4, \sqrt{14})$, $(4, -\sqrt{14})$, $(-5, \sqrt{5})$, and $(-5, -\sqrt{5})$.

The Solutions: Putting It All Together

So, after all that work, we have four solutions to the system of equations: $(4, \sqrt{14})$, $(4, -\sqrt{14})$, $(-5, \sqrt{5})$, and $(-5, -\sqrt{5})$. These are the points where the circle and the parabola intersect. To be absolutely sure, you can always substitute these values back into the original equations to check if they hold true. For example, let's test the solution $(4, \sqrt{14})$: In the first equation, $x^2 + y^2 = 30$, we have $4^2 + (\sqrt{14})^2 = 16 + 14 = 30$. It works! In the second equation, $x - y^2 = -10$, we have $4 - (\sqrt{14})^2 = 4 - 14 = -10$. It also works! You can verify the other solutions the same way. The ability to verify the solutions is a useful skill. This is a very important step to ensure the integrity of the result. Thus, we have successfully solved the system of equations. Congratulations, guys! You did it!

Conclusion: Mastering the Art of Solving

And that's a wrap! We've successfully solved the system of equations. You have now learned how to solve a system of equations by substitution. This is a powerful technique, and with practice, you'll become more and more comfortable with it. Remember that solving systems of equations is a fundamental skill in mathematics and has applications in various fields. Keep practicing, and don't be afraid to try different approaches. The more you work through problems, the better you'll become at recognizing the most efficient methods and the more confident you'll feel in your ability to solve them. You can also explore other methods, such as elimination, to solve systems of equations. Experiment with different strategies to find the one that resonates best with you. Good luck, and keep up the great work, everyone! You got this!