Problem 7 - Algebra and Trigonometry

Question

\(3-8=\) 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

The solutions are \((x, y) = (-25, 5)\) and \((-25, -5)\).

Step 1: Solve for x in the first equation

The first equation is given by \( x + y^2 = 0 \). Solving for \( x \), we rearrange the equation: \( x = -y^2 \). This expression for \( x \) will be used to substitute into the second equation.

Step 2: Substitute x in the second equation

Substitute \( x = -y^2 \) into the second equation \( 2x + 5y^2 = 75 \). This gives us: \( 2(-y^2) + 5y^2 = 75 \). Simplify the equation to \( -2y^2 + 5y^2 = 75 \), which further simplifies to \( 3y^2 = 75 \).

Step 3: Solve for y

We have the equation \( 3y^2 = 75 \). Divide both sides by 3 to get \( y^2 = 25 \). Taking the square root of both sides, we find \( y = 5 \) or \( y = -5 \).

Step 4: Solve for x using the values of y

For \( y = 5 \), substitute back into \( x = -y^2 \), giving \( x = -(5)^2 = -25 \). Similarly, for \( y = -5 \), we again substitute into \( x = -(-5)^2 = -25 \).

Step 5: Verify solutions

Substitute \( (x, y) = (-25, 5) \) and \( (x, y) = (-25, -5) \) back into the original equations to verify they satisfy both equations. For both pairs, \( x + y^2 = -25 + 25 = 0 \) and \( 2x + 5y^2 = 2(-25) + 5(25) = -50 + 125 = 75 \). Thus, both pairs are solutions.

Key Concepts

Understanding System of Equations

A system of equations consists of two or more equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously.
In the context of our exercise, we have the following system of equations:

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

These equations involve the variables \( x \) and \( y \), and we aim to find values of \( x \) and \( y \) that make both statements true at the same time.
The substitution method, which is a key technique to solve systems of equations, involves solving one equation for a variable and then substituting that expression into the other equation(s). This technique effectively reduces the number of variables, making the system easier to solve.

Solving Quadratic Equations

In the process of solving systems, we sometimes end up needing to solve quadratic equations. A quadratic equation is generally expressed in the form \( ax^2 + bx + c = 0 \). This type of equation can have up to two solutions.
For our specific problem, after substitution we simplify to \( 3y^2 = 75 \). To solve this quadratic equation for \( y \), follow these steps:

• Divide both sides by 3 to isolate \( y^2 \): \( y^2 = 25 \).
• Take the square root of both sides to solve for \( y \): \( y = 5 \) or \( y = -5 \).

Thus, the solutions to the quadratic equation are \( y = 5 \) and \( y = -5 \). These solutions are crucial as they will lead us to find the corresponding \( x \) values when substituted back into the original system of equations.

Verification of Solutions

After finding the solutions for \( x \) and \( y \), it is essential to verify that these solutions satisfy the original system of equations. Verification is crucial because it ensures the solutions are correct and applicable to the real-world situations the system models.
Substitute each solution pair back into both original equations to check their validity:

• For the solution pair \((-25, 5)\): We substitute into the equations:
  • \( x + y^2 = -25 + 25 = 0 \)
  • \(2x + 5y^2 = 2(-25) + 5(25) = -50 + 125 = 75\)
• Now for solution pair \((-25, -5)\): Similarly check:
  • \( x + y^2 = -25 + 25 = 0 \)
  • \(2x + 5y^2 = 2(-25) + 5(25) = -50 + 125 = 75\)

Both sets validate the original equations perfectly. This confirms that \((-25, 5)\) and \((-25, -5)\) are indeed correct solutions.