Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=8 \\ y-x=4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y) = (-2, 2)\).

Step 1: Solve the Linear Equation for y

The system of equations consists of one linear equation and one non-linear equation. Start by solving the linear equation for \( y \). Given: \( y - x = 4 \). Add \( x \) to both sides to isolate \( y \): \( y = x + 4 \).

Step 2: Substitute y in the Non-linear Equation

Substitute the expression for \( y \) from Step 1 into the non-linear equation. Replace \( y \) with \( x + 4 \) in \( x^2 + y^2 = 8 \): \[ x^2 + (x + 4)^2 = 8. \] Expand \( (x + 4)^2 \), resulting in \( x^2 + 8x + 16 \). Hence, the equation becomes: \( x^2 + x^2 + 8x + 16 = 8 \).

Step 3: Simplify and Solve for x

Combine like terms: \( 2x^2 + 8x + 16 = 8 \). Subtract 8 from both sides to simplify: \( 2x^2 + 8x + 8 = 0 \). Divide the entire equation by 2 to make it simpler: \( x^2 + 4x + 4 = 0 \). Factor the quadratic equation: \( (x + 2)^2 = 0 \). Solve for \( x \): \( x = -2 \).

Step 4: Substitute x back to find y

Use the solved value of \( x \) and substitute it back into the equation \( y = x + 4 \) to find \( y \). If \( x = -2 \), then \( y = -2 + 4 = 2 \). Thus, the solution for \( (x, y) \) is \( (-2, 2) \).

Step 5: Verify the Solution

Substitute \( x = -2 \) and \( y = 2 \) back into the original equations to ensure they hold true. Check the non-linear equation: \( (-2)^2 + 2^2 = 4 + 4 = 8 \) which is true. Check the linear equation: \( 2 - (-2) = 4 \) which simplifies to \( 4 = 4 \). Both checks pass, confirming the solution.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find values for these variables that satisfy all the equations simultaneously. In this exercise, we have a system comprising two equations:

• \(x^2 + y^2 = 8\) (a non-linear equation due to the squares)
• \(y - x = 4\) (a linear equation because it doesn't involve any squares or other non-linear terms)

By finding a common pair \((x, y)\) that satisfies both equations, we find the solution to the system. One effective method for solving such systems is the substitution method, particularly when one equation is linear, allowing for simpler manipulation and substitution into the other equation.

Linear Equation

A linear equation involves variables to the first power and will graph as a straight line. In our problem, the linear equation is \(y - x = 4\), which can be rewritten to express one variable in terms of the other, for example:

• **Isolating** \(y\): We add \(x\) to both sides, leading to \(y = x + 4\).

This expression is now ready for substitution into the other equation in the system. Linear equations are straightforward because they simplify the process of substitution in complex systems. By isolating \(y\), we've prepared an expression that can be directly inserted into the non-linear equation.

Non-linear Equation

Non-linear equations contain variables whose exponents are not just 1, and typically graph as curves rather than straight lines. In this system, the non-linear equation is \(x^2 + y^2 = 8\). Substituting the expression for \(y\) from the linear equation creates a new form:

• Replace \(y\) with \(x + 4\): \(x^2 + (x + 4)^2 = 8\).

This transformation helps us focus on solving just for \(x\) by removing the variable \(y\). The process then requires expanding \((x + 4)^2\), combining like terms, and finally simplifying it into a quadratic equation.

Quadratic Equation

A quadratic equation is a type of non-linear equation where the highest exponent of the variable is 2, and it usually takes the form \(ax^2 + bx + c = 0\). From our previous steps, we're left with the equation \(x^2 + 4x + 4 = 0\), which is a simplified quadratic equation.

• **Factoring**: Notice that \(x^2 + 4x + 4\) cleans up to \((x + 2)^2 = 0\).
• **Solving**: This factors neatly, indicating that \(x = -2\) is the solution for \(x\).

Quadratic equations can often be solved by factoring, using the quadratic formula, or by completing the square. Here, the problem is structured to neatly factor, leading to a clean solution. Once \(x\) is found, substitute back to find \(y\), confirming that \((x, y)\) satisfies both original equations.