Problem 37 - College Algebra

Question

For the following exercises, use any method to solve the nonlinear system. $$x^{2}+y^{2}-6 y=7$$ $$\quad\quad x^{2}+y=1$$

Step-by-Step Solution

Answer

The solution for \( y \) is \( y = \frac{7 \pm \sqrt{73}}{2} \). Corresponding \( x \) values should be found numerically.

Step 1: Rewrite Equations

First, we'll rewrite the given system of equations:\[ x^2 + y^2 - 6y = 7 \] and \[ x^2 + y = 1 \] so that we can work more easily with them. Notice the second equation can be rearranged to isolate \( x^2 \): \[ x^2 = 1 - y \]

Step 2: Substitute for \(x^2\)

Now that we have \( x^2 = 1 - y \), substitute this expression into the first equation: \[ 1 - y + y^2 - 6y = 7 \]

Step 3: Simplify the Substituted Equation

Simplify the equation from Step 2: \[ y^2 - 6y + 1 - y = 7 \]Combine the \(y\) terms:\[ y^2 - 7y + 1 = 7 \]Subtract 7 from both sides:\[ y^2 - 7y - 6 = 0 \]

Step 4: Solve the Quadratic Equation

Use the quadratic formula to solve \( y^2 - 7y - 6 = 0 \) where \(a = 1\), \(b = -7\), and \(c = -6\):\[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]\[ y = \frac{7 \pm \sqrt{49 + 24}}{2} \]\[ y = \frac{7 \pm \sqrt{73}}{2} \]

Step 5: Find Corresponding \(x\) Values

We found solutions for \( y \). Use these in \( x^2 = 1 - y \) to find \( x \):1. For \( y = \frac{7 + \sqrt{73}}{2} \): \[ x^2 = 1 - \left(\frac{7 + \sqrt{73}}{2}\right) \]2. For \( y = \frac{7 - \sqrt{73}}{2} \): \[ x^2 = 1 - \left(\frac{7 - \sqrt{73}}{2}\right) \] Solve each of these for \( x \).

Step 6: Solve for \(x\) Numerically

To solve for \( x \) numerically use the values obtained for \( y \): 1. Substitute \( y = \frac{7 + \sqrt{73}}{2} \) into \( x^2 = 1 - y \) and simplify.2. Substitute \( y = \frac{7 - \sqrt{73}}{2} \) into \( x^2 = 1 - y \) and simplify.Finally, obtain the square roots to find the potential \( x \) values corresponding to each value of \( y \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of a quadratic equation, which has the standard form: \[ ax^2 + bx + c = 0 \]When dealing with quadratic equations, the quadratic formula offers a way to find solutions by solving for \( x \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This equation helps resolve problems involving any quadratic equation by identifying when solutions exist and what they are.Key Points:

• The term \( b^2 - 4ac \) inside the square root is called the discriminant and dictates the nature of the roots.
• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it is zero, there is one real double root.
• If the discriminant is negative, the roots are complex and non-real.

In our given exercise, applying the quadratic formula was crucial in determining the possible values for \( y \) by plugging \( a = 1 \), \( b = -7 \), and \( c = -6 \) into the equation.

Substitution Method

The substitution method is an effective strategy for solving systems of equations, especially when one of the equations can easily be solved for one variable. It involves replacing a variable in one equation with an expression derived from another equation.For example, in our exercise:

• We start with two equations: \[ x^2 + y^2 - 6y = 7 \] and \[ x^2 + y = 1 \].
• Rearrange the second equation to isolate \( x^2 \): \[ x^2 = 1 - y \].
• Substitute \( 1 - y \) for \( x^2 \) in the first equation, leading to a quadratic in \( y \).

This approach converts a system of nonlinear equations to a problem that can be tackled using techniques for solving quadratic equations, such as the quadratic formula. It reduces complexity by focusing on one variable at a time, simplifying the path to the solution.

Solving Quadratic Equations

When solving quadratic equations, like the transformed \( y^2 - 7y - 6 = 0 \) from our exercise, several methods can be applied, including factoring, using the quadratic formula, and completing the square. Our exercise uses the quadratic formula due to its wide applicability.Here’s a basic outline of solving quadratic equations:

• Factoring: Break down the equation to factors that multiply to the original quadratic. But it may not always be possible.
• Quadratic Formula: As explained earlier, it provides an algebraic solution when factoring is not straightforward.
• Completing the Square: Rewriting the quadratic in a perfect square form to make it easier to solve.

In our exercise, solving \( y^2 - 7y - 6 = 0 \) using the quadratic formula gave the potential \( y \) values. These values were then used to find corresponding \( x \) values to solve the nonlinear system, completing the solution process.