Problem 37 - College Algebra

Question

Use any method to solve the nonlinear system. $$ \begin{array}{r} x^{2}+y^{2}-6 y=7 \\ x^{2}+y=1 \end{array} $$

Step-by-Step Solution

Answer

The solutions to the system are \((1, 0)\) and \((-1, 0)\).

Step 1: Identify the Equations

We are given a system of two equations: 1. \( x^2 + y^2 - 6y = 7 \) 2. \( x^2 + y = 1 \). We will use substitution to solve this system.

Step 2: Solve Second Equation for y

From the second equation, \( x^2 + y = 1 \), we can express \( y \) in terms of \( x \): \[ y = 1 - x^2 \]

Step 3: Substitute y in the First Equation

Substitute \( y = 1 - x^2 \) into the first equation \( x^2 + y^2 - 6y = 7 \): \[ x^2 + (1-x^2)^2 - 6(1-x^2) = 7 \]

Step 4: Simplify the Substituted Equation

Expand and simplify the equation by substituting \( y \): \[ x^2 + (1 - 2x^2 + x^4) - 6 + 6x^2 = 7 \] Combine terms: \[ x^4 + 5x^2 - 12 = 0 \]

Step 5: Solve the Quadratic-like Equation

The equation \( x^4 + 5x^2 - 12 = 0 \) can be treated as a quadratic in terms of \( z = x^2 \):\[ z^2 + 5z - 12 = 0 \]Using the quadratic formula, \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = 5, c = -12 \): \[ z = \frac{-5 \pm \sqrt{25 + 48}}{2} \] \[ z = \frac{-5 \pm 7}{2} \]This gives \( z = 1 \) or \( z = -6 \). Since \( z = x^2 \), the valid solution is \( x^2 = 1 \).

Step 6: Find x-values

From \( x^2 = 1 \), we have two possible values for \( x \):\[ x = 1 \] or \[ x = -1 \]

Step 7: Calculate Corresponding y-values

Using \( y = 1 - x^2 \), for each \( x \): - If \( x = 1 \), then \( y = 1 - 1^2 = 0 \).- If \( x = -1 \), then \( y = 1 - (-1)^2 = 0 \).

Step 8: Verify the Solutions

Check both solutions \((x, y) = (1, 0)\) and \((-1, 0)\) in the original system of equations:- For \((1, 0)\), both equations hold.- For \((-1, 0)\), both equations also hold.Thus, both are valid solutions.

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations. This technique is particularly useful when dealing with nonlinear systems. The idea behind substitution is to express one variable in terms of the other, based on any equation from the system, and then replace this expression in the other equation(s).

• Start by solving one of the equations for one variable.
• Substitute this expression into the other equation(s).
• Solve the new equation, which will now contain only one variable.
• Use the solution to find the value of the other variable by back-substitution.

In our problem, we expressed \( y \) from the equation \( x^2+y=1 \), then substituted it into the first equation. This helped us eliminate \( y \) and focus only on \( x \). This process often simplifies complex systems and leads to finding solutions systematically.

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides the roots of the equation directly, using the coefficients of the terms.The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• \( a \), \( b \), and \( c \) are constants from the quadratic equation.
• The expression \( b^2 - 4ac \) is called the discriminant.
• If the discriminant is positive, there are two distinct real roots.
• A zero discriminant indicates exactly one real root (repeated).
• A negative discriminant leads to complex roots.

In this exercise, after substitution, we formed a quadratic-like equation in terms of \( z = x^2 \). Applying the quadratic formula, we found potential values for \( z \), from which only real and non-negative results gave practical solutions for \( x \).

Solving Quadratic Equations

Solving quadratic equations involves finding the values of the variable that satisfy the equation. There are several methods for solving quadratic equations, which include:

• Factoring
• Completing the square
• Using the quadratic formula
• Graphing

In this exercise, the approach taken was to use the quadratic formula, as the equation was not easily factorable. Once we adapted our equation to a quadratic form in terms of \( z = x^2 \), using the quadratic formula enabled us to file down our potential solutions.Recognizing the structure of equations and knowing which method to apply is key to efficiently solving quadratic equations. This flexibility enhances problem-solving skills, especially in algebra.

Algebraic Equations

Algebraic equations form a broad category of equations involving variables that undergo operations like addition, subtraction, multiplication, division, and exponentiation.

• Linear equations are the simplest form, with variables of power one.
• Quadratic equations involve variables of power two.
• Higher-order polynomials come into play with variables of power three or higher.

The particular system of equations in our exercise is nonlinear, combining terms with \( x^2 \) and \( y^2 \). Through algebraic manipulation using techniques such as substitution and simplification, we converted it into a manageable form to solve.Understanding how to manipulate and transition between different types of algebraic equations is crucial in math. This skill set allows us to handle everything from simple to complex systems elegantly.