Solve System Of Equations: Y^2+x^2=53, Y-x=5

by Kenji Nakamura 45 views

Hey guys! Today, we're diving into the exciting world of solving systems of equations. Specifically, we're tackling a system with one linear and one non-linear equation. Don't worry, it's not as intimidating as it sounds! We'll break it down step by step, making sure you understand every twist and turn. Let's get started!

The Challenge: Our System of Equations

Our mission, should we choose to accept it (and we do!), is to find a solution to the following system of equations:

y^2 + x^2 = 53
y - x = 5

We've got a circle equation (y^2 + x^2 = 53) and a linear equation (y - x = 5). Our job is to find the points (x, y) that satisfy both equations simultaneously. Think of it like finding where a line intersects a circle. There might be two intersection points, one, or none. Let's find out!

Strategy Time: Substitution is Our Friend

When dealing with a mix of linear and non-linear equations, the substitution method is often our best friend. It allows us to express one variable in terms of the other and simplify things. In this case, the second equation, y - x = 5, looks perfect for isolating y.

Step 1: Isolate a Variable

Let's isolate y in the second equation. Adding x to both sides gives us:

y = x + 5

Great! Now we have y expressed in terms of x. This is our key to unlocking the solution.

Step 2: Substitute and Conquer

Now comes the fun part: substitution! We'll replace y in the first equation (y^2 + x^2 = 53) with our new expression (x + 5). This transforms the equation into:

(x + 5)^2 + x^2 = 53

See what we did there? We've eliminated y and now have a single equation with just x. This is a quadratic equation in disguise, ready to be solved.

Step 3: Expand and Simplify

Let's expand the (x + 5)^2 term. Remember the formula (a + b)^2 = a^2 + 2ab + b^2? Applying it here, we get:

(x^2 + 10x + 25) + x^2 = 53

Now, let's combine like terms and rearrange the equation to get it into standard quadratic form (ax^2 + bx + c = 0):

2x^2 + 10x + 25 = 53
2x^2 + 10x - 28 = 0

We've got a quadratic equation! Things are looking good.

Step 4: Solve the Quadratic Equation

We have a few options for solving this quadratic equation: factoring, completing the square, or the quadratic formula. Factoring is often the quickest if it's possible. Let's see if we can factor our equation 2x^2 + 10x - 28 = 0. First, notice that all the coefficients are even, so we can divide the entire equation by 2 to simplify:

x^2 + 5x - 14 = 0

Now, we're looking for two numbers that multiply to -14 and add up to 5. Those numbers are 7 and -2. So, we can factor the quadratic as:

(x + 7)(x - 2) = 0

Setting each factor equal to zero gives us our solutions for x:

x + 7 = 0  =>  x = -7
x - 2 = 0  =>  x = 2

We've found two possible x values: -7 and 2.

Step 5: Find the Corresponding y Values

Remember, we're looking for pairs of (x, y) that satisfy the system. We've got our x values, so now we need to find the corresponding y values. We can use the equation we found earlier, y = x + 5, to do this.

For x = -7:

y = -7 + 5 = -2

So, one solution is (-7, -2).

For x = 2:

y = 2 + 5 = 7

So, another solution is (2, 7).

Step 6: Check Our Solutions (Very Important!)

It's crucial to check our solutions in both original equations to make sure they work. Let's start with (-7, -2):

Equation 1: y^2 + x^2 = 53

(-2)^2 + (-7)^2 = 4 + 49 = 53  (Correct!)

Equation 2: y - x = 5

-2 - (-7) = -2 + 7 = 5  (Correct!)

So, (-7, -2) is definitely a solution.

Now, let's check (2, 7):

Equation 1: y^2 + x^2 = 53

7^2 + 2^2 = 49 + 4 = 53  (Correct!)

Equation 2: y - x = 5

7 - 2 = 5  (Correct!)

(2, 7) is also a solution. Awesome!

Analyzing the Answer Choices

Now that we've solved the system, let's look at the answer choices provided:

A. (-10, -5) B. (-7, -2) C. (7, 2) D. (5, 10)

We found the solutions (-7, -2) and (2, 7). Looking at the choices, B. (-7, -2) is one of our solutions. The correct answer is B.

Key Takeaways for Solving Systems of Equations

  • Substitution is your friend: When you have a mix of linear and non-linear equations, substitution is a powerful technique.
  • Isolate a variable: Choose the easiest variable to isolate in one of the equations.
  • Substitute carefully: Replace the variable in the other equation with the expression you found.
  • Solve the resulting equation: You'll often end up with a quadratic equation, so brush up on your quadratic-solving skills (factoring, quadratic formula, etc.).
  • Find the corresponding values: Once you have one variable, plug it back into one of the equations to find the other variable.
  • Check your solutions: Always, always, always check your solutions in the original equations. This prevents errors and ensures you've found the correct answers.

Mastering the Art of Solving Systems

Solving systems of equations is a fundamental skill in algebra and beyond. The more you practice, the better you'll become at recognizing the best strategies and avoiding common pitfalls. Remember, substitution is just one tool in your toolbox. You'll also encounter situations where elimination or graphing might be more efficient. Keep practicing, and you'll become a system-solving pro!

So there you have it, guys! We've successfully navigated a system of equations, found the solutions, and learned some valuable problem-solving techniques along the way. Keep up the great work, and I'll see you next time with more math adventures!