Solving Systems Of Equations: A Step-by-Step Guide
Hey guys! Today, we're diving into a super important topic in algebra: solving systems of equations. Specifically, we're going to tackle a system where we have a linear equation and a quadratic equation. Don't worry, it sounds scarier than it actually is! We'll break it down step by step, so you'll be solving these like a pro in no time. So, let's jump right into it and figure out how to solve the system of equations:
y = 2x + 4
y = x^2 + x - 2
Understanding Systems of Equations
Before we jump into solving, let's quickly recap what a system of equations actually is. Basically, it's a set of two or more equations that we're trying to solve simultaneously. This means we're looking for the values of the variables (in this case, x and y) that make all the equations in the system true at the same time. Graphically, the solution to a system of equations represents the point(s) where the graphs of the equations intersect. Think of it as finding the common ground between the equations. In our case, we've got a linear equation (y = 2x + 4, which represents a straight line) and a quadratic equation (y = x² + x - 2, which represents a parabola). So, we're looking for the point(s) where the line and the parabola cross each other. This understanding is crucial as we move forward, so make sure you're comfortable with this concept. Remember, we are not just solving one equation but finding the points that satisfy both equations simultaneously.
The Substitution Method: Our Go-To Technique
Okay, so how do we actually solve this thing? One of the most common and effective methods for solving systems of equations is the substitution method. The basic idea behind substitution is to solve one of the equations for one variable and then substitute that expression into the other equation. This eliminates one of the variables, leaving us with a single equation that we can solve for the remaining variable. In our system:
y = 2x + 4
y = x^2 + x - 2
Notice that both equations are already solved for y. This is awesome because it makes our lives much easier! We can directly substitute the expression for y from the first equation (2x + 4) into the second equation. This is the heart of the substitution method – replacing one variable with an equivalent expression to simplify the problem. By doing this, we are essentially merging the two equations into one, which allows us to solve for a single variable. This is a powerful technique, and you'll find it super useful in many different algebraic scenarios. So, let's get to the substitution and see how it unfolds!
Step-by-Step Substitution
Let's take it slow and break down the substitution process. Since we know y = 2x + 4, we can replace the y in the second equation with this expression. This gives us:
2x + 4 = x^2 + x - 2
See what we did there? We've now got a single equation with only one variable (x). This is a huge step forward! Our next goal is to rearrange this equation into a standard form that we can easily solve. Specifically, we want to get it into the form of a quadratic equation: ax² + bx + c = 0. This form is super helpful because we have several methods for solving quadratic equations, like factoring, completing the square, or the quadratic formula. The key here is to manipulate the equation by adding or subtracting terms from both sides until we have zero on one side and the quadratic expression on the other. This process of rearranging the equation is crucial for setting up the next steps in our solution. So, let's move terms around and get this equation into its quadratic form!
Rearranging into Quadratic Form
To get our equation into the standard quadratic form, we need to move all the terms to one side. Let's subtract 2x and 4 from both sides of the equation:
2x + 4 - 2x - 4 = x^2 + x - 2 - 2x - 4
This simplifies to:
0 = x^2 - x - 6
Now we have a quadratic equation in the form ax² + bx + c = 0, where a = 1, b = -1, and c = -6. Awesome! We're one step closer to finding the values of x. Now that we have the equation in this standard form, we can choose our preferred method for solving it. In this case, factoring looks like a pretty straightforward option. Factoring involves breaking down the quadratic expression into two binomial expressions that multiply together to give us the original quadratic. It's like reverse distribution! If factoring doesn't jump out at you, don't worry; we could also use the quadratic formula. But let's try factoring first, as it's often the quickest route if it works. So, let's see if we can factor this quadratic and find our x values!
Factoring the Quadratic Equation
Now, let's factor the quadratic equation x² - x - 6 = 0. We need to find two numbers that multiply to -6 and add up to -1 (the coefficient of our x term). After a little thought, we can see that the numbers -3 and 2 fit the bill perfectly (-3 * 2 = -6 and -3 + 2 = -1). So, we can factor the quadratic as follows:
(x - 3)(x + 2) = 0
This is a fantastic step! We've successfully factored the quadratic, which means we're on the home stretch for finding our x values. The next key idea is the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This property is the cornerstone of solving factored equations, as it allows us to break down a single equation into multiple simpler equations. In our case, it means that either (x - 3) must equal zero or (x + 2) must equal zero (or both!). This gives us two separate equations that we can solve individually for x. So, let's apply the zero product property and find those crucial x values!
Finding the Values of x
Using the zero product property, we set each factor equal to zero:
x - 3 = 0 or x + 2 = 0
Solving these simple equations, we get:
x = 3 or x = -2
Alright! We've found our x values! We know that x can be either 3 or -2. But remember, we're solving a system of equations, which means we need to find the corresponding y values for each of these x values. Each x value will give us a different point of intersection between the line and the parabola. To find the y values, we'll simply substitute each x value back into one of our original equations. It doesn't matter which equation we choose, but the linear equation (y = 2x + 4) is usually easier to work with. So, let's plug in our x values and find the y values that complete our solutions!
Calculating the Corresponding y Values
Now, let's find the y values that correspond to our x values. We'll use the equation y = 2x + 4.
For x = 3:
y = 2(3) + 4 = 6 + 4 = 10
So, when x = 3, y = 10. This gives us one solution: (3, 10).
For x = -2:
y = 2(-2) + 4 = -4 + 4 = 0
So, when x = -2, y = 0. This gives us another solution: (-2, 0).
Excellent work! We've successfully found the y values that correspond to our x values. This means we have the complete solutions to our system of equations. Remember, each solution represents a point where the graphs of the two equations intersect. We found two such points in this case, which is perfectly normal when dealing with a linear and a quadratic equation. Now that we have both solutions, let's take a moment to summarize our findings and make sure we've answered the original question completely.
The Solutions
Therefore, the solutions to the system of equations are:
- (3, 10)
- (-2, 0)
So, the correct answer is D. (-2,0) and (3,10). We did it! We successfully solved the system of equations using the substitution method. We took it step by step, from understanding the problem to finding the x values, calculating the y values, and arriving at our final solutions. This process might seem like a lot of steps, but with practice, it becomes second nature. Remember, the key is to break down the problem into smaller, manageable steps and to understand the logic behind each step. Now that we've solved this particular system, you'll be much better equipped to tackle other systems of equations. Great job, guys!
Key Takeaways and Tips
Before we wrap up, let's highlight some key takeaways and tips that will help you solve systems of equations like a champ:
- The Substitution Method is Your Friend: This method is incredibly versatile and works well for many types of systems, especially when one equation is already solved for a variable or can be easily solved. Remember to substitute the entire expression, not just a single term.
- Rearrange to Quadratic Form: When you end up with a quadratic equation, getting it into the standard form (ax² + bx + c = 0) is crucial. This allows you to use factoring, the quadratic formula, or completing the square to solve for the variable.
- Master Factoring: Factoring is often the quickest way to solve quadratic equations, so practice your factoring skills! Look for two numbers that multiply to the constant term (c) and add up to the coefficient of the x term (b).
- The Zero Product Property is Key: Don't forget this powerful property! If the product of two factors is zero, at least one of the factors must be zero. This allows you to split a factored equation into two simpler equations.
- Don't Forget the y Values! Once you've found the x values, remember to substitute them back into one of the original equations to find the corresponding y values. This gives you the complete solutions as ordered pairs.
- Check Your Answers: It's always a good idea to check your solutions by plugging them back into both original equations. If both equations are true for your solutions, you know you've got it right!
By keeping these tips in mind and practicing regularly, you'll become a system-solving superstar! Solving systems of equations is a fundamental skill in algebra and has many applications in real-world scenarios. So, keep practicing, keep learning, and you'll be amazed at what you can achieve!