Problem 20 - Precalculus: Mathematics 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: Analyze the System of Equations

We need to solve the system of equations composed of two equations: \(x - y^2 = 0\) and \(y - x^2 = 0\). Our goal is to find values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Express Variables in Terms of Each Other

From the first equation, we have \(x = y^2\). Substituting this into the second equation gives us \(y - (y^2)^2 = 0\). This simplifies to \(y - y^4 = 0\) or \(y(1 - y^3) = 0\).

Step 3: Solve for y

The equation \(y(1 - y^3) = 0\) gives us \(y = 0\) or \(1 - y^3 = 0\). Solving \(1 - y^3 = 0\) leads to \(y^3 = 1\), hence \(y = 1\). Therefore, the possible values for \(y\) are \(0\) and \(1\).

Step 4: Solve for x Using the y Solutions

Substitute \(y = 0\) into \(x = y^2\) to find \(x = 0^2 = 0\). Substitute \(y = 1\) into \(x = y^2\) to find \(x = 1^2 = 1\).

Step 5: Write the Solutions

The solutions to the system are \((x, y) = (0, 0)\) and \((x, y) = (1, 1)\). These pairs satisfy both original equations.

Key Concepts

Solution of Equations

Finding the solution of equations involves identifying values that satisfy each equation in a system. In simple terms, you want to find numbers that can replace the variables and make the equation true.
In the given system, we have two equations:

• \(x - y^2 = 0\)
• \(y - x^2 = 0\)

We are looking for values of \(x\) and \(y\) that solve, or satisfy, both equations at the same time. This requires either substituting one equation into the other or manipulating the equations to isolate one of the variables.
Solving a system of equations can seem tricky, but with practice, it becomes easier to see the necessary steps. Start by expressing one variable in terms of the other. This allows us to substitute and simplify until we can find potential values for the variables. Finally, we test these potential solutions to ensure they satisfy both equations.

Simultaneous Equations

Simultaneous equations are equations that must be satisfied together. We need to find a common solution that works for both at the same time. In this exercise, the two equations were:

• \(x = y^2\)
• \(y = x^2\)

This means, we need values for \(x\) and \(y\) that make both these equations true simultaneously.
To solve simultaneous equations, we often use methods like substitution, elimination, or graphical approaches. This example used substitution by expressing \(x\) in terms of \(y\) from the first equation and substituting it into the second equation. Such methods are effective because they reduce the system to solve a simpler problem, one that involves fewer variables or simpler expressions.
Once the potential solutions are found, it's crucial to check them against both original equations. Even though a value might work for one, it must work for both to be a true solution to the system.

Non-Linear Equations

Non-linear equations are those in which variables are raised to a power other than one. They can be challenging because they do not form straight lines when graphed. In our exercise, both equations, \(x = y^2\) and \(y = x^2\), are non-linear because they involve squares of the variables.
Solving non-linear equations often requires different strategies compared to linear ones. For instance, these types of equations can have multiple solutions. Therefore, it’s crucial to explore all potential options when solving them, as done by considering both \(y = 0\) and \(y = 1\) in the workout.
It’s helpful to remember that non-linear systems often result in multiple possible solution pairs. After we solved using algebra, the solutions we found were \((0, 0)\) and \((1, 1)\), as these points satisfy both non-linear equations in this system. Recognizing the characteristics of non-linear equations helps understand why certain solutions are valid and how they are derived.