Problem 7 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the System of Equations

The given system of equations is:1. \( x + y^2 = 0 \)2. \( 2x + 5y^2 = 75 \)

Step 2: Express One Variable in Terms of Another

From equation 1, solve for \( x \):\( x = -y^2 \).This expression allows us to substitute \( x \) in the second equation.

Step 3: Substitute into the Second Equation

Substitute \( x = -y^2 \) into equation 2:\( 2(-y^2) + 5y^2 = 75 \).

Step 4: Simplify the Substituted Equation

Simplify the equation:\(-2y^2 + 5y^2 = 75\), which reduces to \(3y^2 = 75\).

Step 5: Solve for the Variable

Solve for \( y^2 \):\( y^2 = \frac{75}{3} \),which simplifies to \( y^2 = 25 \).Taking the square root, we get two solutions: \( y = 5 \) and \( y = -5 \).

Step 6: Substitute Back to Solve for x

Use \( x = -y^2 \) to find corresponding \( x \) values:1. If \( y = 5 \), then \( x = -(5)^2 = -25 \).2. If \( y = -5 \), then \( x = -(-5)^2 = -25 \).

Step 7: Verify Solutions in Original Equations

Verify \((x, y) = (-25, 5)\) and \((-25, -5)\) in original equations:1. \( x + y^2 = 0 \) holds as \( -25 + 25 = 0 \).2. \( 2x + 5y^2 = 75 \) holds as \( 2(-25) + 5(25) = -50 + 125 = 75 \).Both solutions satisfy the original system.

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations with the same variables. Solving these systems allows us to find values for the variables that satisfy all equations simultaneously. In this exercise, the system involves two equations with variables \(x\) and \(y\):

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

The goal is to find all pairs \((x, y)\) that satisfy both equations.
There are several methods to solve such systems, including graphing, elimination, and substitution. Here, we'll focus on the substitution method. This involves expressing one variable in terms of another and substituting this expression into the second equation. This simplifies the system to a single equation in one variable, making it easier to solve.

Solution Verification

After solving a system of equations, it's crucial to verify that the solutions satisfy both original equations. This step ensures accuracy and confirms that no mistakes were made during calculation. For our solutions \((-25, 5)\) and \((-25, -5)\), we need to check:
1. **For \((-25, 5)\):** - In the first equation: \( x + y^2 = 0 \), becoming \(-25 + 25 = 0\), which holds true. - In the second equation: \( 2x + 5y^2 = 75 \), becoming \( -50 + 125 = 75\), which is also valid.2. **For \((-25, -5)\):** - Checking the first equation: \(-25 + 25 = 0\) is satisfied. - Checking the second equation: \( -50 + 125 = 75\) is confirmed correct.Both solutions satisfy the system, confirming they are indeed the correct solutions. This process is essential for solution integrity, especially for complex systems.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, generally in the form \( ax^2 + bx + c = 0 \). In our exercise, we encounter quadratic elements when simplifying the substituted system. After substitution, we derived:

• \(3y^2 = 75\)

Solving for \(y^2\) by dividing both sides by 3, gives:

• \( y^2 = 25 \).

Taking the square root provides two possible solutions for \(y\):

• \( y = 5 \)
• \( y = -5 \)

Quadratic equations like this can have up to two solutions because the square root function can yield both positive and negative roots.
This mathematical property is critical for finding all solutions to a given quadratic equation, which is why it plays a significant role in solving systems involving quadratic relations.