Solving Systems Of Equations: Find A Solution!

by Andrew McMorgan 47 views

Hey math enthusiasts! Today, we're diving into the exciting world of solving systems of equations. Specifically, we're tackling this problem: What is a solution to the system of equations: y - 3 = x and x^2 - 6x + 13 = y? We'll break down the problem step-by-step, making sure everyone, from math newbies to seasoned pros, can follow along. So, grab your pencils, and let's get started!

Understanding the Problem: A System of Equations

First things first, let's understand what a system of equations actually is. A system of equations is simply a set of two or more equations that share the same variables. In our case, we have two equations, both involving the variables 'x' and 'y'. The goal here is to find the values of 'x' and 'y' that satisfy both equations simultaneously. Think of it like finding the perfect meeting point for two different lines or curves on a graph. This meeting point, represented as a coordinate pair (x, y), is the solution we're after. So, before we even think about diving into the nitty-gritty of algebra, let's visualize what we're trying to achieve. We want a pair of numbers that, when plugged into both equations, make both equations true. It's like finding the key that unlocks two doors at once!

To find the solution, we need to find the values of x and y that make both equations true at the same time. The given system of equations is:

  1. y - 3 = x
  2. x² - 6x + 13 = y

We are also given four possible solutions:

  • (-5, 2)
  • (-2, 1)
  • (2, 5)
  • (8, 5)

Our mission, should we choose to accept it (and we do!), is to figure out which of these pairs of numbers, if any, actually works when we plug them into our equations. It might seem daunting at first, but don't worry, we're going to take it one step at a time. Think of it as a detective game where we're searching for the correct combination. Each possible solution is a suspect, and we need to test their alibis to see if they hold up under scrutiny. So, let's put on our detective hats and get ready to investigate!

Method 1: Substitution – A Classic Approach

One common method for solving systems of equations is substitution. The idea behind substitution is to solve one equation for one variable and then substitute that expression into the other equation. This way, we reduce the system to a single equation with a single variable, which is much easier to solve. In our case, the first equation, y - 3 = x, is already nicely set up for substitution. We can easily isolate 'x' in this equation. It's like having one piece of the puzzle already in place, which gives us a great head start.

From the first equation, we have: x = y - 3

Now, we can substitute this expression for 'x' into the second equation: (y - 3)² - 6(y - 3) + 13 = y

See what we did there? We replaced 'x' in the second equation with the expression 'y - 3'. This is the heart of the substitution method. By doing this, we've transformed the second equation from an equation with both 'x' and 'y' into an equation with only 'y'. Now we have a single equation that we can solve for 'y'. It's like turning a complex problem into a simpler one, which is a powerful technique in mathematics. Now, let's roll up our sleeves and get ready to do some algebra to simplify and solve this equation for 'y'.

Let’s simplify and solve for 'y':

(y² - 6y + 9) - 6y + 18 + 13 = y y² - 6y + 9 - 6y + 18 + 13 - y = 0 y² - 13y + 40 = 0

Now we have a quadratic equation in terms of 'y'. Time to put our factoring skills to the test! Quadratic equations are equations of the form ay² + by + c = 0, and they appear quite frequently in math problems. Solving them is a crucial skill, so let's make sure we're comfortable with the process. There are several ways to solve a quadratic equation, including factoring, using the quadratic formula, or completing the square. In this case, the equation factors nicely, which is always a bonus.

Let’s factor the quadratic equation: (y - 5)(y - 8) = 0

From this factored form, we can see that the possible values for 'y' are: y = 5 or y = 8

So, we've found two potential values for 'y'! That's a big step forward. But remember, we're looking for a pair of numbers (x, y) that satisfy both equations. We've found the 'y' values, but we still need to find the corresponding 'x' values. This is where we go back to our original equations and use the 'y' values we just found to calculate the 'x' values. It's like we've solved half the puzzle, and now we need to find the matching pieces to complete the picture. Let's see what 'x' values correspond to y = 5 and y = 8.

Now, let’s find the corresponding values for 'x' using the equation x = y - 3:

If y = 5, then x = 5 - 3 = 2 If y = 8, then x = 8 - 3 = 5

So, we have two potential solutions:

  • (2, 5)
  • (5, 8)

But wait! We're not done yet. We've found two potential solutions, but we need to make sure they actually work. This is a crucial step in solving any system of equations. We need to plug each potential solution back into the original equations and see if they satisfy both equations. It's like testing a key in a lock to make sure it actually opens the door. We don't want to jump to conclusions and assume we have the right answer until we've verified it. So, let's put these solutions to the test!

Method 2: Testing the Given Options – A Practical Shortcut

Sometimes, the most efficient way to solve a problem is to use the information provided. In this case, we are given four possible solutions. Instead of solving the system from scratch, we can test each of these options to see which one satisfies both equations. This can be a much faster approach, especially if the solutions are relatively simple. It's like having a multiple-choice test where you can try out each option to see which one fits the question best. Let's put this method into action!

Let's test each option in both equations:

  • Option 1: (-5, 2)

    • Equation 1: 2 - 3 = -5 (True)
    • Equation 2: (-5)² - 6(-5) + 13 = 25 + 30 + 13 = 68 ≠ 2 (False)
    • So, (-5, 2) is not a solution.

  • Option 2: (-2, 1)

    • Equation 1: 1 - 3 = -2 (True)
    • Equation 2: (-2)² - 6(-2) + 13 = 4 + 12 + 13 = 29 ≠ 1 (False)
    • So, (-2, 1) is not a solution.

  • Option 3: (2, 5)

    • Equation 1: 5 - 3 = 2 (True)
    • Equation 2: (2)² - 6(2) + 13 = 4 - 12 + 13 = 5 (True)
    • So, (2, 5) is a solution!

  • Option 4: (8, 5)

    • Equation 1: 5 - 3 = 8 (False)
    • So, (8, 5) is not a solution.

As we can see, the option (2, 5) satisfies both equations, making it the correct solution. This method highlights the importance of utilizing all the information given in a problem. Sometimes, the answer is right in front of you, and all you need to do is test it out! It's like having a shortcut on a map – why take the long route when you can get there faster?

Verifying the Solution – Double-Checking Our Work

It's always a good practice to verify your solutions to ensure accuracy. Even if we're confident in our calculations, a small error can sometimes slip in. To verify a solution, we simply plug the values of x and y back into the original equations and check if they hold true. This is like proofreading an essay – you want to make sure everything is perfect before you submit it. In our case, we've already done this step as part of Method 2, but let's reiterate the process for emphasis.

Let’s verify the solution (2, 5):

  • Equation 1: y - 3 = x
    • 5 - 3 = 2 (True)
  • Equation 2: x² - 6x + 13 = y
    • (2)² - 6(2) + 13 = 4 - 12 + 13 = 5 (True)

Since (2, 5) satisfies both equations, we can confidently say that it is indeed a solution to the system. This step of verification is crucial, especially in exams or when dealing with complex problems. It's a way to catch any mistakes and ensure that you're submitting the correct answer. Think of it as the final seal of approval on your work!

Conclusion: The Power of Systems of Equations

So, there you have it! We've successfully solved the system of equations and found that (2, 5) is a solution. We explored two different methods: substitution and testing the given options. Both methods led us to the same answer, highlighting the versatility of mathematical problem-solving. Remember, there's often more than one way to crack a math problem, and it's important to find the method that works best for you.

Systems of equations are a fundamental concept in mathematics and have wide-ranging applications in various fields, from engineering and physics to economics and computer science. They allow us to model and solve real-world problems involving multiple variables and relationships. Mastering the techniques for solving systems of equations is a valuable skill that will serve you well in your mathematical journey and beyond.

Hopefully, this step-by-step guide has made the process of solving systems of equations clearer and more approachable. Keep practicing, and you'll become a pro in no time! And remember, math can be fun, especially when you're cracking a tough problem. Keep exploring, keep learning, and keep enjoying the beauty of mathematics!