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: Analyzing Given Equations

We have a system 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: Solving the First Equation for x

The first equation is \( x - y^2 = 0 \). Solving for \( x \), we find \( x = y^2 \). This means that for any solution, \( x \) must be equal to the square of \( y \).

Step 3: Substituting into the Second Equation

Substitute \( x = y^2 \) into the second equation \( y - x^2 = 0 \). This gives us \( y - (y^2)^2 = 0 \), simplifying to \( y = y^4 \).

Step 4: Solving the Simplified Equation

Now solve the equation \( y = y^4 \). Arrange it as \( y^4 - y = 0 \), factor it to \( y(y^3 - 1) = 0 \). So, the solutions are \( y = 0 \) or \( y^3 = 1 \).

Step 5: Finding Values of y

Solve \( y^3 = 1 \). The real root is \( y = 1 \). Therefore, the values of \( y \) are \( y = 0 \) and \( y = 1 \).

Step 6: Finding Corresponding x Values

Using \( x = y^2 \), find the corresponding \( x \) values:- If \( y = 0 \), then \( x = 0^2 = 0 \).- If \( y = 1 \), then \( x = 1^2 = 1 \).

Step 7: Forming Solution Pairs

The solution pairs \((x, y)\) that satisfy both equations are:- \((0, 0)\)- \((1, 1)\)

Key Concepts

Simultaneous Equations

Simultaneous equations involve solving two or more equations at once to find common values that satisfy all given equations. These types of problems are critical because they often model real-world situations where different conditions need to be considered together.
For example, this particular exercise asks us to solve two equations simultaneously:

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

The objective is to find values of \( x \) and \( y \) that satisfy both equations. This means finding a set of points \((x, y)\) that lie on the curves represented by these equations at the same time. Such points are called solution pairs, and our goal is to determine which pairs exist.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, often taking the form \( ax^2 + bx + c = 0 \). In solving our system of equations, we encounter quadratic forms:

• The term \( x = y^2 \) from the first equation represents a parabola, which is a typical graph of a quadratic equation.
• The substitution of \( x = y^2 \) into the second equation results in \( y = y^4 \), leading to \( y^4 - y = 0 \).

By factoring \( y^4 - y = 0 \) into \( y(y^3 - 1) = 0 \), we further simplify the problem into finding solutions for each factor, ensuring our solutions meet the criteria of a quadratic form. Recognizing quadratic characteristics helps in restructuring and solving these equations effectively.

Solution Pairs

When dealing with systems of equations, finding solution pairs is the ultimate goal. These are pairs of values \((x, y)\) that satisfy both equations in the system. For the problem at hand, once we solve the system, we must determine what \( y \) values solve the factored equation:

• \( y = 0 \)
• \( y^3 = 1 \)

The real-valued solution to \( y^3 = 1 \) is \( y = 1 \).
Then, we substitute these solutions back to find the corresponding \( x \) values using \( x = y^2 \). This yields the solution pairs:

• \((0, 0)\) when both \( y = 0 \) and \( x = 0 \)
• \((1, 1)\) when both \( y = 1 \) and \( x = 1 \)

These solution pairs ensure both original equations are satisfied, illustrating how the simultaneous system can be resolved clearly to meet both conditions.