Problem 16 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{array}{l} x^{2}+y^{2}=16 \\ y+2 x=-1 \end{array}\right.$$

Step-by-Step Solution

Answer

Solutions: \( x = \frac{-4 \pm \sqrt{316}}{10} \), corresponding \( y \).

Step 1: Solve for y in the second equation

The second equation is given as \( y + 2x = -1 \). We can solve this equation for \( y \) by isolating \( y \) on one side of the equation. Subtract \( 2x \) from both sides to get:\[ y = -1 - 2x \]

Step 2: Substitute y in the first equation

Now that we have \( y \) in terms of \( x \), substitute \( y = -1 - 2x \) into the first equation \( x^2 + y^2 = 16 \). This gives us:\[ x^2 + (-1 - 2x)^2 = 16 \]

Step 3: Expand the equation

Expand the equation from the substitution so we can simplify and solve for \( x \):\[ (-1 - 2x)^2 = (-1)^2 + 2(-1)(2x) + (2x)^2 = 1 + 4x + 4x^2\]Thus, the equation becomes:\[ x^2 + 1 + 4x + 4x^2 = 16 \]

Step 4: Simplify the equation

Combine like terms to simplify the equation to a standard quadratic form:\[ 5x^2 + 4x + 1 = 16 \]Subtract 16 from both sides to set the equation to zero:\[ 5x^2 + 4x - 15 = 0 \]

Step 5: Solve the quadratic equation

We use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = 4 \), and \( c = -15 \). First, calculate the discriminant:\[ b^2 - 4ac = 4^2 - 4 \times 5 \times (-15) = 16 + 300 = 316 \]Calculate \( x \):\[ x = \frac{-4 \pm \sqrt{316}}{10} \]Simplifying gives two solutions for \( x \): \( x = \frac{-4 + \sqrt{316}}{10} \) and \( x = \frac{-4 - \sqrt{316}}{10} \)

Step 6: Solve for y using x values

Using the values of \( x \) from Step 5, substitute back into \( y = -1 - 2x \) to find the corresponding \( y \) values.\[ y_1 = -1 - 2 \times \frac{-4 + \sqrt{316}}{10} \] and \[ y_2 = -1 - 2 \times \frac{-4 - \sqrt{316}}{10} \]Calculate the numeric values to find these pairs (\( x_1, y_1 \)) and (\( x_2, y_2 \)).

Key Concepts

The Substitution Method

The substitution method is a popular and effective technique used to solve systems of equations. This method involves expressing one of the variables in terms of the other and then substituting it into another equation. Our exercise provides a straightforward example.
To begin, look at the pair of equations given:

• Main equation: \( x^2 + y^2 = 16 \)
• Sub-equation: \( y + 2x = -1 \)

The second equation can be solved for \( y \), making it \( y = -1 - 2x \). This expression of \( y \) is then inserted into the main equation.
By doing this, we transform the two-variable problem into a one-variable problem, simplifying our work dramatically. This method is efficient and, when applied systematically, can reliably yield the solutions to complex systems of equations.

Understanding Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). These equations can have two solutions, one solution, or sometimes no solution.
In the given exercise, after substitution and simplifying, we arrive at the quadratic equation \( 5x^2 + 4x - 15 = 0 \).
Solving quadratic equations often involves using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here:

• \( a = 5 \)
• \( b = 4 \)
• \( c = -15 \)

These are coefficients derived from rearranging and simplifying the substitution into our main equation.
Through understanding and applying the properties of quadratic equations, including the discriminant \( b^2 - 4ac \), we can ascertain the nature of the solutions, ensuring that we capture all possible values of \( x \).

The Art of Solving Equations

Solving equations is about strategic manipulation and exploration to uncover values that satisfy given conditions. Here, step-by-step methods improve efficiency and accuracy.
Once substitution condenses the system into a solvable equation, follow procedural operations:

• Expand the terms when necessary. Use distribution laws, such as \((a + b)^2 = a^2 + 2ab + b^2\).
• Simplify by combining like terms. This results in an easy-to-read equation that is less prone to errors.
• Isolate variables. It reduces clutter and focuses on one entity at a time.

Each step has its importance, incrementally bringing you closer to an exact solution. Whether working manually or computationally, clear articulation of each step is crucial in solving equations.

Algebraic Manipulation

Algebraic manipulation techniques are pivotal in adjusting and solving equations. They help simplify, rearrange, and isolate variables to make equations more manageable.

• Start by rearranging to make one variable expressed concerning another, as was done in expressing \( y = -1 - 2x \).
• Use algebraic operations like addition, subtraction, multiplication, or division to maintain the equation's balance. This ensures that any changes still hold true.
• Factor and simplify expressions where applicable, converting complex polynomials into simpler forms.

Skilled manipulation allows for efficient solving and ensures equations retain their integrity as transformations occur. As seen in our exercise, such methods facilitate the solution's pathway, allowing for accurate and consistent results.