Problem 69 - A Graphical Approach to Precalculus with Limits

Question

Solve each nonlinear system of equations analytically. $$\begin{aligned}&y=-x^{2}+2\\\&x-y=0\end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \((-2, -2)\) and \((1, 1)\).

Step 1: Express One Variable in Terms of the Other

The second equation in the system is \( x - y = 0 \). We can solve this equation for \( y \) to express it in terms of \( x \). Add \( y \) to both sides to get \( x = y \), which means that \( y = x \).

Step 2: Substitute the Expression into the First Equation

Now that we know \( y = x \), substitute this into the first equation \( y = -x^2 + 2 \). Replace the \( y \) in this equation with \( x \), resulting in the equation \( x = -x^2 + 2 \).

Step 3: Rearrange the Equation to Find x

The equation \( x = -x^2 + 2 \) can be rearranged to find \( x \). Add \( x^2 \) to both sides to get \( x^2 + x = 2 \).

Step 4: Solve the Quadratic Equation

The equation \( x^2 + x - 2 = 0 \) is a quadratic equation in standard form. Factorize it as \((x + 2)(x - 1) = 0\). Set each factor to zero and solve for \( x \):- \( x + 2 = 0 \) gives \( x = -2 \)- \( x - 1 = 0 \) gives \( x = 1 \)

Step 5: Find Corresponding y-values

Using the relation \( y = x \), find the corresponding values of \( y \) for each solution of \( x \):- If \( x = -2 \), then \( y = -2 \).- If \( x = 1 \), then \( y = 1 \).

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra that represents any equation in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations are essential in solving many real-world problems due to their pattern of depicting parabolic shapes. In our exercise, we encountered the equation \( x^2 + x - 2 = 0 \), which is in the standard quadratic form.
Quadratic equations can be solved using several methods, including factorization, the quadratic formula, or completing the square. Each method provides a systematic approach to finding values for \( x \) that satisfy the equation. The solutions to these equations are often referred to as roots or zeros of the equation, and they can be real or complex numbers depending on the discriminant \( b^2 - 4ac \).
Understanding quadratic equations is crucial because they form the basis for more complex algebraic equations and appear frequently in scientific and engineering calculations.

Substitution Method

The substitution method is a valuable technique for solving systems of equations, particularly when one equation can easily be rewritten to express one variable in terms of the other. This method was effectively used in our exercise involving the equations \( y = -x^2 + 2 \) and \( x - y = 0 \).
By rewriting \( x - y = 0 \) as \( y = x \), we can insert this expression into the first equation. This substitution simplifies the problem from a system of equations to a single equation in one variable. The resulting equation \( x = -x^2 + 2 \) is more straightforward to manage and solve.
This method helps in systematically breaking down complex problems into manageable parts, allowing for easier manipulation and solution of nonlinear systems of equations. It highlights the effectiveness of isolating variables as a strategic approach in algebra.

Factorization

Factorization is a powerful tool for solving quadratic equations, which involves expressing the equation as a product of its linear factors. In this exercise, we used factorization to solve the quadratic equation \( x^2 + x - 2 = 0 \).
This process began by identifying two numbers that multiply to give the constant term \(-2\), while also adding up to the coefficient of \( x \), which is \(1\). The numbers \(2\) and \(-1\) satisfy these requirements, allowing us to rewrite the equation as \( (x + 2)(x - 1) = 0 \).
Factorizing the equation simplifies our task by providing two linear equations to solve: \( x + 2 = 0 \) and \( x - 1 = 0 \). Solving each gives us the potential solutions for \( x \). This technique is efficient, especially when the quadratic can be easily decomposed into factors, facilitating the discovery of solutions quickly.

Analytical Solution

An analytical solution is a meticulous approach to solving equations, where we obtain exact answers through algebraic formulations and reasoning, as opposed to numerical methods or approximations.
In this nonlinear system, we applied analytical methods to derive solutions for \( x \) and \( y \) that satisfy both equations concurrently. Using techniques like substitution and factorization, we arrived at precise solutions: \( x = -2 \) with \( y = -2 \), and \( x = 1 \) with \( y = 1 \).
The advantage of analytical solutions lies in their exactness and reliability. They provide clear insight into the relationships between variables and help verify the correctness of derived solutions. Mastering this approach equips students with the skills to tackle any algebraic problem confidently.