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

Hey there, math enthusiasts! Today, we're going to dive into the exciting world of solving systems of equations. Specifically, we'll tackle the following system:

$\left\{\begin{array}{l} y=-x^2+2 x+1 \\ y=x-1 \end{array}\right.$

Don't worry if it looks a little intimidating at first; we'll break it down into manageable steps, making it easy for you to understand. This is a great example because it combines a quadratic equation (the one with the x squared) and a linear equation (the one with just x). Let's get started!

Understanding Systems of Equations

Before we jump into the solution, let's quickly review what a system of equations actually is. Simply put, it's a set of two or more equations that we want to solve simultaneously. This means we're looking for the values of x and y that satisfy all the equations in the system. The solution(s) to a system of equations represent the point(s) where the graphs of the equations intersect. In our case, one equation is a parabola (because it's quadratic), and the other is a straight line. The solutions will be the points where the line crosses the parabola.

There are several ways to solve a system of equations, and the best method depends on the specific equations. Common methods include substitution, elimination, and graphing. In this particular case, substitution is a very efficient approach. The goal is to find values for the variables that satisfy all equations in the system. The solutions represent the points where the graphs of the equations intersect. For a quadratic and a linear equation, there can be zero, one, or two solutions. So, in our example, we are looking for the points where the parabola and the line intersect. If they don't intersect, there is no solution; if they touch at one point, there is one solution; and if they cross twice, there are two solutions. Keep this in mind as we proceed!

The Substitution Method: Our Weapon of Choice

For our system of equations, the substitution method is the most straightforward. The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. In this case, both equations are already solved for y, which makes our job even easier! We can substitute the expression for y from the second equation (y = x - 1) into the first equation.

Let's get down to business. Since we know that y equals both -x² + 2x + 1 and x - 1, we can set these two expressions equal to each other:

x - 1 = -x² + 2x + 1

See? We've now got a single equation with only one variable, x. This is a much easier problem to solve. Our focus is now to manipulate this equation to solve for x. Remember, our ultimate goal is to find the values of x and y that satisfy both original equations. So let's get solving!

Solving for x: The Heart of the Matter

Now that we've set up our equation, let's solve for x. We'll start by rearranging the equation to get all terms on one side and set it equal to zero. This is a standard approach when solving quadratic equations.

x - 1 = -x² + 2x + 1

Add x² to both sides: x² + x - 1 = 2x + 1

Subtract 2x from both sides: x² - x - 1 = 1

Subtract 1 from both sides: x² - x - 2 = 0

Now we have a quadratic equation in the standard form ax² + bx + c = 0. We can solve this by factoring, completing the square, or using the quadratic formula. Let's try factoring.

We need to find two numbers that multiply to -2 (the constant term) and add up to -1 (the coefficient of the x term). Those numbers are -2 and 1. So, we can factor the quadratic equation as follows:

(x - 2)(x + 1) = 0

For this equation to be true, either (x - 2) = 0 or (x + 1) = 0. Therefore:

x - 2 = 0 => x = 2 x + 1 = 0 => x = -1

We have found two possible values for x: 2 and -1. Remember, x and y represent the coordinates of the points where the line and the parabola intersect. Now we need to find the corresponding y values for each of these x values.

Finding the y Values: Completing the Picture

Now that we have our x values, it's time to find the corresponding y values. We can use either of the original equations to do this. The second equation, y = x - 1, is the easiest to use. Let's substitute each x value we found into this equation:

For x = 2: y = 2 - 1 y = 1

For x = -1: y = -1 - 1 y = -2

So, we have two points of intersection: (2, 1) and (-1, -2). These are the solutions to our system of equations!

Verifying Our Solution: Always a Good Idea

Before we celebrate, it's always a good idea to verify our solution. We can plug the (x, y) coordinates of our solutions back into both original equations to make sure they work.

Let's check the point (2, 1):

Equation 1: y = -x² + 2x + 1 1 = -(2)² + 2(2) + 1 1 = -4 + 4 + 1 1 = 1 (This checks out!)

Equation 2: y = x - 1 1 = 2 - 1 1 = 1 (This also checks out!)

Now let's check the point (-1, -2):

Equation 1: y = -x² + 2x + 1 -2 = -(-1)² + 2(-1) + 1 -2 = -1 - 2 + 1 -2 = -2 (This checks out!)

Equation 2: y = x - 1 -2 = -1 - 1 -2 = -2 (This also checks out!)

Both points satisfy both equations, so our solutions are correct!

Conclusion: You Did It!

Congratulations! You've successfully solved a system of equations involving a quadratic and a linear equation. We started with a set of equations, used the substitution method to simplify, solved for x, found the corresponding y values, and verified our solution. This process is a fundamental skill in algebra, and with practice, you'll become a pro at solving these types of problems. Remember to always double-check your work to ensure accuracy.

Solving systems of equations is a valuable skill in many areas of mathematics and science. The ability to find the intersection points of different functions can be used in numerous real-world applications, from calculating the optimal mix of resources to predicting the trajectory of a projectile. Keep practicing, and you'll find that these problems become easier and more enjoyable to solve. The more you practice, the more comfortable you will become with different types of equations and different solution methods.

Further Exploration

Want to dig deeper? Here are some related topics you might find interesting:

• Different methods for solving systems of equations (elimination, graphing). This offers you more options, depending on the equations.
• Systems of inequalities. Similar to systems of equations, but you are dealing with inequalities, leading to solutions represented by regions, not just points.
• Applications of systems of equations in real-world scenarios. Learning the practical uses can help motivate your learning.
• More complex systems of equations, including those involving other types of functions.

By exploring these topics, you'll continue to build your math skills and understanding. Math can be fun and rewarding, and with the right approach, anyone can master it. Keep exploring and asking questions. Math is all around us, and the more you learn, the more you will appreciate its beauty and power.

Remember, practice is key. Try solving similar problems on your own, and don't be afraid to make mistakes. Mistakes are a natural part of the learning process. Each time you solve a problem, you solidify your understanding and increase your confidence.

For additional practice and in-depth explanations, check out resources like Khan Academy or your textbook.

If you enjoyed this, you might be interested in resources that expand on algebra concepts, or resources focused on more complex math ideas, like Calculus. Solving systems of equations is a critical step in these more advanced mathematical pursuits.

For more practice and a deeper understanding, I recommend you check out Khan Academy, a great resource for learning math.

Khan Academy - Systems of Equations