Solve System Of Equations: Find X Values
Hey guys! Let's dive into solving a system of equations today. We've got a fun one that mixes a circle equation with a linear equation. It's like finding where a line crashes into a circle – super cool, right? We're going to break it down step-by-step, so you'll be a pro at this in no time. Our main goal here is to figure out the x values that make both equations true. So, let's get started and unravel this mathematical puzzle!
System of Equations
Alright, let's kick things off by stating the system of equations we're going to tackle. We have:
This is a classic setup where we have a circle equation (x² + y² = 25) and a linear equation (x + y = 1). The circle equation represents a circle centered at the origin (0, 0) with a radius of 5, while the linear equation represents a straight line. Our mission, should we choose to accept it (and we totally do!), is to find the points (x, y) where these two graphs intersect. These points of intersection are the solutions to the system. We're specifically hunting for the x values of these points, which will tell us the x-coordinates where the line and circle meet. To do this, we're going to use a method called substitution, which is a slick way of solving systems of equations. Basically, we'll solve one equation for one variable and then plug that expression into the other equation. This will give us a single equation in one variable, which we can then solve. It might sound a bit complicated now, but trust me, it's like riding a bike – once you get the hang of it, you'll be cruising through these problems. So, let's roll up our sleeves and get into the nitty-gritty of solving this system!
Solving for y in the Linear Equation
Okay, first things first, let's grab that linear equation, x + y = 1, because it looks simpler to handle. Our immediate goal here is to isolate y on one side of the equation. Think of it as getting y its own VIP space so we can see what it's equal to in terms of x. This is a crucial step because once we have y expressed in terms of x, we can substitute that expression into the circle equation. It's like having a secret code that unlocks the solution! To isolate y, we need to get rid of that pesky x term on the left side. The way we do that is by subtracting x from both sides of the equation. Remember, whatever we do to one side, we have to do to the other to keep the equation balanced. It's like a mathematical seesaw – we want to keep it perfectly level. So, when we subtract x from both sides, we get: y = 1 - x. There you have it! We've successfully solved for y. Now we know that y is equal to 1 minus x. This is our golden ticket, the expression we're going to use to substitute into the circle equation. This substitution is going to transform our problem into something we can solve much more easily. We're taking a big step towards finding those x values, so let's keep the momentum going!
Substituting into the Circle Equation
Now comes the fun part – substitution! We've got y = 1 - x, and we're going to take this expression and plug it into the circle equation, x² + y² = 25. Think of it like replacing y in the circle equation with its equivalent value in terms of x. This is a slick move because it transforms our two-variable equation into a single-variable equation, which is way easier to solve. When we substitute, we get x² + (1 - x)² = 25. Notice how we've replaced y with (1 - x)? Now, we have an equation that only involves x, which means we're on the home stretch to finding those x values. But, hold your horses! We're not done yet. We need to simplify this equation before we can solve it. This means we need to expand that (1 - x)² term. Remember, (1 - x)² is the same as (1 - x) multiplied by itself. So, we'll need to use the distributive property (or the FOIL method, if you're into that) to expand it. Once we've expanded and simplified, we'll have a quadratic equation in x. And guess what? We know how to solve those! We can use factoring, the quadratic formula, or even completing the square. So, buckle up, because we're about to dive into some algebraic acrobatics to simplify this equation and get one step closer to our solution. Let's do this!
Expanding and Simplifying
Alright, let's roll up our sleeves and get down to the nitty-gritty of expanding and simplifying the equation we landed on: x² + (1 - x)² = 25. First up, we need to tackle that (1 - x)² term. Remember, this means (1 - x) multiplied by (1 - x). To expand this, we'll use the distributive property, which is just a fancy way of saying we'll multiply each term in the first set of parentheses by each term in the second set. So, (1 - x)(1 - x) becomes 11 - 1x - x1 + x * x, which simplifies to 1 - 2x + x². Now, we can substitute this back into our equation, giving us x² + 1 - 2x + x² = 25. We're not done yet, though! We need to combine like terms to make our equation look even cleaner. We have two x² terms, which combine to give us 2x². So, our equation now looks like 2x² - 2x + 1 = 25. Looking good, right? But we're not quite there yet. To solve this quadratic equation, we need to set it equal to zero. This means we need to subtract 25 from both sides of the equation. When we do that, we get 2x² - 2x - 24 = 0. Bingo! We now have a quadratic equation in standard form, which is super exciting because we have all sorts of tools to solve these. We could use the quadratic formula, factoring, or completing the square. But before we jump into any of those methods, let's see if we can simplify our equation even further. Notice that all the coefficients (2, -2, and -24) are divisible by 2? That means we can divide the entire equation by 2 to make the numbers smaller and easier to work with. When we divide by 2, we get x² - x - 12 = 0. Awesome! We've transformed our original equation into a much simpler quadratic equation. Now we're ready to roll into solving for x. Let's do it!
Factoring the Quadratic Equation
Alright, we've got a shiny, simplified quadratic equation: x² - x - 12 = 0. Now it's time to unleash our factoring skills! Factoring is like reverse-engineering multiplication – we're trying to find two binomials (expressions with two terms) that, when multiplied together, give us our quadratic equation. To factor x² - x - 12, we need to find two numbers that multiply to -12 (the constant term) and add up to -1 (the coefficient of the x term). Think of it like a puzzle – we're searching for the perfect pieces that fit together. Let's brainstorm some factors of -12: we've got 1 and -12, -1 and 12, 2 and -6, -2 and 6, 3 and -4, and -3 and 4. Which pair adds up to -1? Ding ding ding! It's 3 and -4. So, we can rewrite our quadratic equation in factored form as (x + 3)(x - 4) = 0. This is a major breakthrough! We've transformed our quadratic equation into a product of two binomials that equals zero. Now, here's the magic trick: if the product of two things is zero, then at least one of those things must be zero. This is called the zero-product property, and it's a key concept in solving factored equations. It means that either (x + 3) = 0 or (x - 4) = 0 (or both!). So, we've broken our problem down into two simpler equations, each of which we can solve easily. We're in the final stretch now. All we need to do is solve each of these mini-equations to find our x values. Let's go for it!
Solving for x
Okay, we're in the home stretch! We've factored our quadratic equation into (x + 3)(x - 4) = 0, and we know that this means either x + 3 = 0 or x - 4 = 0. Let's tackle the first equation, x + 3 = 0. To solve for x, we need to isolate it on one side of the equation. We can do this by subtracting 3 from both sides. This gives us x = -3. Awesome! We've found one of our x values. Now, let's move on to the second equation, x - 4 = 0. Again, we want to isolate x, so we'll add 4 to both sides of the equation. This gives us x = 4. Fantastic! We've found our second x value. So, the x values that satisfy our quadratic equation are x = -3 and x = 4. But wait, we're not quite done yet! Remember, we started with a system of equations, and we've only found the x values. To be super thorough, we could plug these x values back into either of our original equations to find the corresponding y values. This would give us the full coordinates of the points where the line and circle intersect. However, the question specifically asks for the x values, so we've technically answered it. We've successfully navigated the twists and turns of this problem, from substitution to factoring to solving. Give yourselves a pat on the back – you've earned it! We've found the values of x that make the system of equations true.
Final Answer
Alright, we've journeyed through the land of equations, battled algebraic beasts, and emerged victorious! We started with a system of equations, used substitution to transform it, factored a quadratic equation, and finally, arrived at our destination: the x values that satisfy the system. We found that x = -3 and x = 4. So, the final answer is that the values of x for which (x, y) will be a solution of the system of equations are x = -3 and x = 4. You guys nailed it! Solving systems of equations can feel like a puzzle, but with the right tools and a bit of practice, you can conquer any mathematical challenge. Keep up the awesome work, and remember, math is not just about numbers and equations – it's about problem-solving and critical thinking, skills that will serve you well in all areas of life. So, keep exploring, keep questioning, and keep pushing your mathematical boundaries. You've got this!
Therefore, the correct answer is D. and
