Solving Quadratic Systems: A Step-by-Step Guide

Hey guys! Let's dive into the exciting world of solving systems of equations algebraically, especially when quadratics are involved. In this guide, we'll tackle a specific problem and break down each step so you can confidently solve similar equations. We'll focus on making the process super clear and easy to follow, ensuring you understand not just how to get the answer, but why it works.

The Challenge: A System with a Quadratic

Our mission, should we choose to accept it (and we do!), is to solve this system of equations:

$$\begin{cases} -10 = 11x + x^2 \\ -12x = 30 \end{cases}$$

We've already got one solution: (-4, -18). Our goal? To find the other one. Buckle up, because we're about to embark on an algebraic adventure!

Step 1: Deciphering the Linear Equation

First things first, let's zero in on the linear equation: -12x = 30. This equation is straightforward, and our aim is to isolate 'x'. How do we do it? By dividing both sides by -12. Let's do the math:

$$x = \frac{30}{-12} = -\frac{5}{2}$$

So, we've discovered that x = -5/2. This is a crucial piece of the puzzle. Now we know the x-coordinate of our second solution.

Step 2: Plugging into the Quadratic Equation

Now that we know x = -5/2, let's take that value and substitute it into the quadratic equation: -10 = 11x + x^2. This step is all about finding the corresponding 'y' value for our 'x'.

Let's plug it in:

$$-10 = 11(-\frac{5}{2}) + (-\frac{5}{2})^2$$

Time to simplify. First, we calculate the terms:

$$-10 = -\frac{55}{2} + \frac{25}{4}$$

To combine these, we need a common denominator. Let's use 4:

$$-10 = -\frac{110}{4} + \frac{25}{4}$$

Now we can add the fractions:

$$-10 = -\frac{85}{4}$$

Step 3: Isolating 'y' (The Twist!)

Wait a minute! There's a slight misunderstanding here. The original problem presents the first equation as a relationship between x and a constant (-10), not a system where we need to explicitly solve for 'y'. What we've done by substituting x = -5/2 into the quadratic equation is to check if this x-value satisfies the equation. It turns out that:

$$-10 \ne -\frac{85}{4}$$

This tells us that x = -5/2 does not satisfy the quadratic equation. So, there seems to be a mix-up in how the system is interpreted. The problem implies the system should yield pairs (x, y), but the first equation doesn't explicitly involve 'y'.

However, if we were to assume there's a 'y' implicitly linked and the system intended to find (x, y) pairs that satisfy both equations, we need to rethink our approach slightly. The second equation, -12x = 30, only gives us the x-value. Without a second equation that involves 'y', we can't directly solve for 'y'.

Let's reframe the problem:

Assuming the intention was to find x-values that satisfy the quadratic equation, and the solutions provided (-4, -18) hints at a misunderstanding of the system's nature, let's focus on solving the quadratic equation itself.

Step 4: Solving the Quadratic Equation

Let's rewrite our quadratic equation in the standard form: x² + 11x + 10 = 0. This makes it easier to solve.

We can solve this by factoring. We're looking for two numbers that multiply to 10 and add up to 11. Those numbers are 10 and 1. So, we can factor the equation as:

$$(x + 10)(x + 1) = 0$$

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

$$x + 10 = 0 \Rightarrow x = -10$$

$$x + 1 = 0 \Rightarrow x = -1$$

So, the x-values that satisfy the quadratic equation are x = -10 and x = -1.

Step 5: Finding the Corresponding 'y' Values (Revised Understanding)

Given the second equation -12x = 30, we already found x = -5/2. If the problem intended a system where this x must also fit some implicit 'y' relationship, and since the quadratic doesn't define 'y', there's a disconnect.

However, if we interpret the system as two separate conditions:

1. The quadratic equation x² + 11x + 10 = 0
2. The linear equation -12x = 30

We solved them independently. The quadratic gives us x = -10 and x = -1. The linear gives us x = -5/2. There's no single (x, y) that satisfies both in a conventional system of equations sense because there's no 'y' in the quadratic equation.

The provided solution (-4, -18) seems to fit a different system entirely, possibly one where 'y' is explicitly defined in both equations.

Therefore, based on the equations given, and acknowledging the likely misunderstanding in the problem's setup, we've clarified the solutions for 'x' for each equation independently.

Conclusion: Untangling the Equations

Alright, guys, we've taken a deep dive into this system of equations, and it turned out to be a bit of a puzzle! We navigated through the algebra, solved the quadratic by factoring, and pinpointed the x-values that make the equations true. While the initial problem setup seemed to hint at a standard system of equations, we discovered that the equations are essentially independent conditions on 'x'.

We found the x-values that satisfy the quadratic equation (x = -10 and x = -1) and the x-value that satisfies the linear equation (x = -5/2). The key takeaway here is to carefully analyze the equations and understand what they're asking before jumping into the solution. Keep practicing, and you'll become an equation-solving pro in no time!