Solving Y = X² + 1 & 5x + 3y = 25: A Deep Dive

by Admin 47 views
Solving y = x² + 1 & 5x + 3y = 25: A Deep Dive

Hey guys! Let's dive into a classic algebra problem: solving a system of equations where we have a quadratic equation (y = x² + 1) and a linear equation (5x + 3y = 25). This is super common, and understanding how to solve these is a key skill. We'll break down the steps, explain the logic, and make sure you're comfortable with the process. Get ready to flex those math muscles! We will learn how to approach the problem, step by step, which includes understanding the two equations separately, solving for a single variable, substituting this value to find the others and finally understanding the solution.

Understanding the Equations

First off, let's take a good look at our equations individually. The first one, y = x² + 1, is a quadratic equation. This means if you were to graph it, you'd get a parabola – a U-shaped curve. The '+ 1' shifts the parabola upwards by one unit on the y-axis, making the vertex (the bottom point of the U) at (0, 1). Remember, with a quadratic equation, we're expecting either zero, one, or two solutions (points where the parabola intersects with the line). The second equation, 5x + 3y = 25, is a linear equation. When graphed, this gives us a straight line. We can rearrange this equation to make it easier to understand by solving for y.

Let's do that quickly: Subtract 5x from both sides: 3y = -5x + 25. Then, divide everything by 3: y = (-5/3)x + 25/3. Now it's in slope-intercept form (y = mx + b), where 'm' is the slope (-5/3) and 'b' is the y-intercept (25/3). The slope tells us how steep the line is, and the y-intercept tells us where the line crosses the y-axis. Knowing this, we know the line slopes downwards from left to right. Solving this type of problem helps in understanding the relationship between linear and quadratic equations. It helps us visualize how these equations interact with each other when graphed, and where their solution is, if there is one. We can find the solutions to these equations by substitution. The intersection of the two graphs gives us the solution(s) to our problem. We will now learn how to solve these problems by substitution, which is one of the most common methods to find the solution.

Understanding the components of the question is really important so you know what you are looking for. You are essentially looking for where the graphs intersect, which are the solutions. Having the correct solutions is what is required, so that will be our final goal. Also, make sure that you practice as much as you can. It really helps when you do this, because you will understand how the equations work and how you can work with the solution.

Solving by Substitution: The Key Method

Alright, now for the fun part: solving the system! The most straightforward method here is substitution. Since we already have y isolated in our first equation (y = x² + 1), we can substitute this expression for y in our second equation. This gives us a single equation with only x as the variable, which we can then solve. Let's do it step by step:

  1. Substitute: Replace y in the second equation (5x + 3y = 25) with (x² + 1): 5x + 3(x² + 1) = 25.
  2. Simplify: Distribute the 3: 5x + 3x² + 3 = 25.
  3. Rearrange: Get everything on one side to set the equation to zero. Subtract 25 from both sides and rearrange to get the standard quadratic form (ax² + bx + c = 0): 3x² + 5x - 22 = 0.

Now we have a quadratic equation to solve. There are a couple of ways we can do this: factoring, completing the square, or using the quadratic formula. Let's explore the methods and choose one. Remember, you can always check your answers by graphing the equations and seeing where they intersect, which helps you visualize the solutions. The methods we will discuss can be applied to many other similar problems. But you must also understand the basic elements of the question, or you won't be able to solve the problem at all. Let's start with factoring to find the roots and complete the question.

Solving the Quadratic Equation: Factoring

Let's try to solve the quadratic equation 3x² + 5x - 22 = 0 by factoring. Factoring can be a quick way to solve a quadratic, but it doesn't always work easily. We are looking for two numbers that multiply to give us (3 * -22 = -66) and add up to 5. After some thought, we find that 11 and -6 work. So, we can rewrite the middle term of our quadratic equation: 3x² + 11x - 6x - 22 = 0.

Now, we can factor by grouping: Factor out an x from the first two terms and a -2 from the last two terms: x(3x + 11) - 2(3x + 11) = 0.

Notice that we have a common factor of (3x + 11). We can factor this out: (3x + 11)(x - 2) = 0.

For the product of two factors to be zero, at least one of them must be zero. So, we set each factor equal to zero and solve for x:

  • 3x + 11 = 0 => 3x = -11 => x = -11/3
  • x - 2 = 0 => x = 2

We have found two possible x values: x = -11/3 and x = 2. Now we need to find the corresponding y values for each of these. We can substitute these values back into either of our original equations, but the first one (y = x² + 1) is easier to work with. Remember that the quadratic equation has, in general, two solutions and this is what we expected and found. Let's go ahead and find the solution.

Finding the y-values and the Solution

Now that we have our x-values, we can plug them back into the equation y = x² + 1 to find the corresponding y values.

For x = -11/3: y = (-11/3)² + 1 = 121/9 + 1 = 121/9 + 9/9 = 130/9.

So, one solution is (-11/3, 130/9).

For x = 2: y = 2² + 1 = 4 + 1 = 5.

So, another solution is (2, 5).

Therefore, the solutions to the system of equations are (-11/3, 130/9) and (2, 5). These are the points where the parabola and the line intersect. Congratulations, guys, we solved it! The most important aspect is to know how the equations are formed and how you can work to find the answers. You also need to understand how each equation works. In general, make sure you know the rules and the way to follow them, and always use examples to solidify your knowledge. Practicing is also very important, because you will solidify your knowledge with each passing question.

Checking Your Answer: Always a Good Idea

It's always a good practice to check your answers! You can do this by plugging the x and y values of your solutions back into both of the original equations to make sure they are true. For example:

Check the solution (2, 5):

  • Equation 1 (y = x² + 1): 5 = 2² + 1 => 5 = 4 + 1 => 5 = 5 (True!)
  • Equation 2 (5x + 3y = 25): 5(2) + 3(5) = 25 => 10 + 15 = 25 => 25 = 25 (True!)

Check the solution (-11/3, 130/9):

  • Equation 1 (y = x² + 1): 130/9 = (-11/3)² + 1 => 130/9 = 121/9 + 9/9 => 130/9 = 130/9 (True!)
  • Equation 2 (5x + 3y = 25): 5(-11/3) + 3(130/9) = 25 => -55/3 + 130/3 = 25 => 75/3 = 25 => 25 = 25 (True!)

Both solutions satisfy both equations, so we know we've done it correctly. Feel proud of yourselves, guys. You successfully solved a system of equations involving a quadratic! Always double-check your work, and you will be good to go. This whole process will become easier the more you practice. Make sure you practice every day and you will see how it gets easier. So go for it, guys! The more you practice, the more you will know. Now you will learn how to approach more complicated problems.

Alternative Methods: Quadratic Formula

If you couldn't factor the quadratic equation, don't sweat it! You can always use the quadratic formula to find the solutions to the quadratic equation (3x² + 5x - 22 = 0). The quadratic formula is: x = (-b ± √(b² - 4ac)) / 2a, where a = 3, b = 5, and c = -22. So, let's plug these values into the formula and see what happens.

x = (-5 ± √(5² - 4 * 3 * -22)) / (2 * 3) x = (-5 ± √(25 + 264)) / 6 x = (-5 ± √289) / 6 x = (-5 ± 17) / 6

This gives us two solutions:

  • x = (-5 + 17) / 6 = 12 / 6 = 2
  • x = (-5 - 17) / 6 = -22 / 6 = -11/3

As you can see, we get the same x-values we found through factoring. This just shows that there is more than one way to skin a cat. Use the method you are more comfortable with or the one that works best for the problem. You can always check with the other method. No matter the method you use, the result must be correct. Make sure that you understand the rules of the methods you will use and that you follow them. After you get more experience, these methods will come naturally, and you will be able to solve them with ease.

Key Takeaways

  • Substitution is a powerful tool: It allows you to solve systems of equations by eliminating one variable.
  • Know your quadratic equations: Recognize the form (ax² + bx + c = 0) and be familiar with methods like factoring and the quadratic formula.
  • Always check your answers: Plugging your solutions back into the original equations is the best way to ensure accuracy.
  • Practice makes perfect: The more you solve these types of problems, the more confident and efficient you'll become.

Congratulations, guys! You've successfully navigated the process of solving a system of equations involving a quadratic and a linear equation. Keep practicing, keep learning, and keep asking questions. Mathematics is all about exploration, and it's a lot more fun when you understand the 'why' behind the 'how'. Keep up the great work, and happy solving!