Problem 7 - Algebra and Trigonometry

Question

Use the substitution method to find all solutions of the system of equations. \(\left\\{\begin{aligned} x+y^{2} &=0 \\ 2 x+5 y^{2} &=75 \end{aligned}\right.\)

Step-by-Step Solution

Answer

Solutions: \((-25, 5)\) and \((-25, -5)\).

Step 1: Solve First Equation for x

The first equation is \( x + y^2 = 0 \). To express \( x \) in terms of \( y \), subtract \( y^2 \) from both sides, giving \( x = -y^2 \).

Step 2: Substitute Expression for x into Second Equation

Substitute \( x = -y^2 \) into the second equation: \( 2(-y^2) + 5y^2 = 75 \). This simplifies to \( -2y^2 + 5y^2 = 75 \).

Step 3: Simplify and Solve for y

Combine the terms in the equation: \( 3y^2 = 75 \). Divide both sides by 3 to isolate \( y^2 \): \( y^2 = 25 \). Taking the square root of both sides gives \( y = 5 \) or \( y = -5 \).

Step 4: Find Corresponding Values for x

Use each \( y \) value in the expression \( x = -y^2 \) to find \( x \). For \( y = 5 \), \( x = -(5)^2 = -25 \). For \( y = -5 \), \( x = -(-5)^2 = -25 \).

Step 5: Write Final Solutions as Ordered Pairs

The solutions as ordered pairs are \((-25, 5)\) and \((-25, -5)\).

Key Concepts

System of Equations

In mathematics, a system of equations is a collection of two or more equations with the same set of variables. When solving such a system, the goal is to find values for the variables that satisfy all the equations simultaneously. Systems can consist of linear or nonlinear equations. In our exercise, we deal with a system of two equations involving two variables: \(x\) and \(y\). This system is also a mix of linear and nonlinear, as seen in the equations:

• \(x + y^2 = 0\)
• \(2x + 5y^2 = 75\)

These equations must hold true at the same time, meaning we look for common solutions. The substitution method is one effective approach to solving such systems, working well especially when one of the variables can be easily isolated.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. It's the unifying thread of almost all mathematics and is crucial when working with systems of equations.
In algebra, we use variables (like \(x\) and \(y\)) to represent unknown numbers. These variables can then be manipulated using operations like addition, subtraction, multiplication, and division to form an equation or a system of equations.
In this specific exercise, the application of algebra is evident when we isolate the variable \(x\) in terms of \(y\) from the first equation:

• \(x = -y^2\)

This result is obtained by performing algebraic operations: subtracting \(y^2\) from both sides of the equation \(x + y^2 = 0\). By mastering these operations, you can rearrange and solve equations effectively.

Solving Equations

Solving equations involves finding the value(s) of the variable(s) that satisfy the equation. In our case, the solution process involves a few important steps, especially using the substitution method. Let's break this down into manageable parts:

• First, take an equation where one variable can be easily expressed in terms of the other, like we did by getting \(x = -y^2\).
• Then, substitute this expression into the other equation. This allows you to work with just one variable, simplifying the task.
• Simplify and solve the resulting equation for the isolated variable. For instance, substituting \(-y^2\) for \(x\) in the second equation gave us \(3y^2 = 75\).
• Next, solve for \(y\) by dividing both sides by 3, giving \(y^2 = 25\), and then take the square root to find possible values of \(y\), which are \(5\) and \(-5\).
• Finally, substitute back these values of \(y\) to find corresponding values of \(x\), achieving the full set of solutions.

Each stage of solving requires careful algebraic manipulation and understanding of mathematical principles to ensure potential solutions are both valid and complete.