Problem 56 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} y=x+1 \\ x^{2}-y^{2}=1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \), \( y = 0 \).

Step 1: Express y from the first equation

We are given the equation \( y = x + 1 \). This equation directly expresses \( y \) in terms of \( x \), so \( y = x + 1 \).

Step 2: Substitute y in the second equation

Substitute \( y = x + 1 \) into the second equation \( x^2 - y^2 = 1 \). This gives \( x^2 - (x + 1)^2 = 1 \).

Step 3: Simplify the equation

Expand \((x + 1)^2\) to get \(x^2 + 2x + 1\). Therefore, the equation becomes \( x^2 - (x^2 + 2x + 1) = 1 \).

Step 4: Solve the resulting equation

Simplify to get \(-2x - 1 = 1\). Then solve for \( x \): \(-2x = 2\), resulting in \( x = -1 \).

Step 5: Find the corresponding y value

Substitute \( x = -1 \) back into \( y = x + 1 \). This gives \( y = -1 + 1 = 0 \).

Step 6: Verify the solution

Plug \( x = -1 \) and \( y = 0 \) back into the second equation: \( (-1)^2 - 0^2 = 1 \). The equality holds, confirming the solution is correct.

Key Concepts

Algebraic Solutions

An algebraic solution refers to finding the values of variables that satisfy an equation or a system of equations. In algebra, equations can be very straightforward or complex, incorporating various expressions. Solving these effectively requires understanding the basic principles:

• Identify the equations in your system.
• Choose a method to solve, such as substitution, elimination, or graphing.
• Manipulate the equations to isolate variables.

In this specific exercise, we began with two equations: one linear and another quadratic. Together, they form a system. The goal was to find values of \(x\) and \(y\) that satisfy both equations at once. By applying algebraic techniques like substitution, you can simplify the problem and systematically solve for the unknowns.

Substitution Method

The substitution method is a powerful technique used to solve systems of equations, especially suitable when one equation is easily solvable for one variable. In our problem, the first equation is already solved for \(y\): \(y = x + 1\). Here's how substitution works:

• Solve one of the equations for one variable, if it's not already done.
• Substitute this expression into the other equation. This transforms a system into a single equation with one variable.
• Solve the new equation for the unknown variable.
• Once one variable is found, substitute back to find the other.

By using substitution, we've turned the system into a single quadratic equation. This allowed us to focus on solving for \(x\), making the problem much simpler.

Quadratic Equations

Quadratic equations play an essential role in algebra. They are polynomials of the form \(ax^2 + bx + c = 0\). In this problem, we closely encountered a quadratic form when substituting \(y = x + 1\) into the second equation, leading to \(x^2 - (x + 1)^2 = 1\). Simplifying gives a classic quadratic expression, even though it quickly simplifies further to a linear one:

• Expand and simplify any squared terms, as shown when expanding \((x + 1)^2\).
• Combine like terms to simplify the equation.
• Use algebraic operations to isolate the variable.
• Determine the solutions, which can be real, complex, or redundant.

In many cases, you will encounter quadratic equations with more than one solution. In this particular problem, simplifying led us to a straightforward solution for \(x\), which we verified to ensure accuracy. Quadratic equations can seem challenging, but with practice, they become more manageable.