Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: 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

Solve for \( x \) using the quadratic formula and then find \( y \) using substitution. The solutions are \( (x_1, y_1) \) and \( (x_2, y_2) \).

Step 1: Express y in terms of x

Start by solving the second equation \( y + 2x = -1 \) for \( y \). Subtract \( 2x \) from both sides to get: \[ y = -1 - 2x \]

Step 2: Substitute y in the first equation

Substitute \( y = -1 - 2x \) into the first equation \( x^2 + y^2 = 16 \):\[ x^2 + (-1 - 2x)^2 = 16 \]

Step 3: Expand the expression

Expand \((-1 - 2x)^2 \) in the equation:\[ x^2 + (1 + 4x + 4x^2) = 16 \]Simplify to:\[ 5x^2 + 4x + 1 = 16 \]

Step 4: Simplify the equation

Subtract \( 16 \) from both sides:\[ 5x^2 + 4x + 1 - 16 = 0 \]Which results in:\[ 5x^2 + 4x - 15 = 0 \]

Step 5: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = 4 \), and \( c = -15 \):Calculate the discriminant:\[ b^2 - 4ac = 4^2 - 4(5)(-15) = 16 + 300 = 316 \]The roots are:\[ x = \frac{-4 \pm \sqrt{316}}{10} \]

Step 6: Calculate the x-values

Calculate the two possible values for \( x \):\( x_1 = \frac{-4 + \sqrt{316}}{10} \) \( x_2 = \frac{-4 - \sqrt{316}}{10} \)

Step 7: Substitute x-values into y equation

Substitute \( x_1 \) and \( x_2 \) back into \( y = -1 - 2x \) to find \( y_1 \) and \( y_2 \):For \( x_1 \):\[ y_1 = -1 - 2\left(\frac{-4 + \sqrt{316}}{10}\right) \]For \( x_2 \):\[ y_2 = -1 - 2\left(\frac{-4 - \sqrt{316}}{10}\right) \]

Step 8: Simplified solution

Calculate the specific values of \( y_1 \) and \( y_2 \) by simplifying the expressions, which will give you the pairs \( (x_1, y_1) \) and \( (x_2, y_2) \).

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves replacing one variable with an expression derived from another equation. This technique is particularly useful when working with a system of two equations. By simplifying the system, it makes finding the solution more manageable.
In our example, we start with the two equations:

• \( x^2 + y^2 = 16 \)
• \( y + 2x = -1 \)

First, solve for one variable in terms of the other using the simpler equation, which is often linear. Here, from the second equation, we solve for \( y \) giving us \( y = -1 - 2x \).
Next, substitute this expression into the first equation. This creates a new equation with just one variable, which can then be solved more easily. This method is systematic and reduces the complexity of solving systems.

Quadratic Equation

Quadratic equations are fundamental in algebra and appear as polynomial equations of degree 2. They are generally in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
In our exercise, after substitution, we arrive at the quadratic equation \( 5x^2 + 4x - 15 = 0 \). Quadratics can often be solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = 4 \), and \( c = -15 \).

• Calculate the discriminant \( b^2 - 4ac = 316 \).
• Since \( 316 > 0 \), there are two distinct real solutions.

These solutions provide us the values for \( x \) which, when substituted back into the expression for \( y \), will help find the entire set of solutions for the system.

Algebraic Substitution

Algebraic substitution transforms a system of equations into a more manageable form. This plays a key role in the substitution method. By expressing one variable in terms of another and substituting it into a different equation, you effectively reduce the number of variables.
In the exercise, we express \( y = -1 - 2x \) and substitute this into \( x^2 + y^2 = 16 \). This eliminates the variable \( y \) from the quadratic equation.
Substitution helps to focus on one variable at a time, simplifying the process. It allows us to use simpler algebraic expressions to manage larger, more complex systems. The strategy here is isolation and substitution, streamlining the path to a solution.

Solving Equations

Solving equations is about finding the values of variables that make the equation true. This is a core aspect of algebra. When you solve a system of equations, like in our exercise, you're seeking the values that satisfy all equations simultaneously.
Begin by manipulating the equations to isolate terms and substitute where necessary, as seen with our substitution approach. Once the substitution yields a quadratic equation, using the quadratic formula is often the way forward.
Calculating the quadratic formula involves:

• Identifying values for \( a \), \( b \), and \( c \).
• Finding the discriminant \( b^2 - 4ac \) to determine nature and number of solutions.
• Solving for \( x \) and substituting back to find corresponding \( y \).

Careful attention to each step ensures the solution is correct, and multiple solutions are investigated thoroughly to fulfill all given equations.