Problem 24 - Precalculus: Mathematical for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l}x-y^{2}=0 \\\y-x^{2}=0\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are (0, 0) and (1, 1).

Step 1: Understand Each Equation

The system consists of two equations: \( x - y^2 = 0 \) and \( y - x^2 = 0 \). In the first equation, \( x = y^2 \), showing that \( x \) is expressed in terms of \( y \). In the second equation, \( y = x^2 \), showing that \( y \) is expressed in terms of \( x \).

Step 2: Substitute One Equation into the Other

Since \( x = y^2 \) from the first equation, substitute this into the second equation \( y = x^2 \). This gives us \( y = (y^2)^2 \), which simplifies to \( y = y^4 \).

Step 3: Solve the Equation \( y = y^4 \)

Rewrite the equation as \( y^4 - y = 0 \). Factor this to get \( y(y^3 - 1) = 0 \). This implies two scenarios: either \( y = 0 \) or \( y^3 = 1 \).

Step 4: Solve Each Scenario for \( y \)

1. If \( y = 0 \), substitute \( y \) back into \( x = y^2 \) to find \( x = 0 \). Thus, one solution is \((0, 0)\).2. If \( y^3 = 1 \), then \( y = 1 \). Substitute back into \( x = y^2 \) to find \( x = 1 \). Thus, a second solution is \((1, 1)\).

Step 5: Conclude the Solutions

From the steps above, the solutions for the system of equations are \((0, 0)\) and \((1, 1)\).

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations by expressing one variable in terms of another and then substituting it into the other equation. This can simplify the problem, allowing you to solve for one variable first and then use the result to find the second variable.

In the given exercise, we start with the equations:

• \( x - y^2 = 0 \)
• \( y - x^2 = 0 \)

The substitution method involves taking one equation and solving for one of the variables. Here, from the first equation, we solve for \( x \) in terms of \( y \):
\[ x = y^2 \]We then substitute \( x = y^2 \) into the second equation, turning it into a one-variable equation. This simplifies the process, as we directly work with:
\[ y = (y^2)^2 \]Now, we can solve for \( y \) alone, which becomes a simpler polynomial equation. Substitution is particularly helpful because it breaks down a complex system into easier parts. Once we find \( y \), we substitute it back to find \( x \). Thus, substitution is a straightforward, systematic method to solve systems of equations.

Factoring Equations

Factoring is a mathematical process of breaking down an equation into a product of simpler components that can be solved individually. Factoring is often used after substitution transforms a system into a single equation.
For the solved exercise, after substituting we get the equation:
\[ y = y^4 \]By rearranging, we have:
\[ y^4 - y = 0 \]To solve this, we use factoring to express it as:
\[ y(y^3 - 1) = 0 \]The factored form yields two potential solutions, each forming a simpler equation to solve:

• \( y = 0 \)
• \( y^3 - 1 = 0 \)

This breaks our problem into smaller parts, making it easy to find \( y \) values that satisfy the original problem. Factoring is crucial because it reveals multiple solutions that might not be evident from the initial form of the equation. It's a powerful tool for simplifying and eventually solving polynomial equations.

Polynomial Equations

Polynomial equations are expressions set in the form \( a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 = 0 \), where the highest power of the variable defines the degree of the polynomial.
Let's consider the determining steps from the exercise. After substitution, we handled the polynomial equation:
\[ y^4 - y = 0 \]The degree of the equation here is four, which means there could be up to four possible solutions for \( y \). Solving this required factorization, resulting in smaller polynomial equations like:

• \( y = 0 \)
• \( y^3 - 1 = 0 \)

For \( y^3 = 1 \), the real root is \( y = 1 \), indicating the cube root of one. Each solution to these polynomial equations corresponds to a possible value for \( y \) that satisfies the system.
Understanding polynomial equations is vital because they often arise when substituting or manipulating the original system. Knowing how to factor and solve them allows us to unlock complex systems and find comprehensive solutions efficiently.