Problem 25 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} &x=\sqrt{y}\\\ &x^{2}-9 y^{2}=9 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions of the given system of equations are: \(\left(x_1, y_1\right) = \left(\sqrt{\frac{1 + \sqrt{325}}{18}}, \frac{1 + \sqrt{325}}{18}\right)\) \(\left(x_2, y_2\right) = \left(\sqrt{\frac{1 - \sqrt{325}}{18}}, \frac{1 - \sqrt{325}}{18}\right)\)

Step 1: Express x in terms of y

Given the first equation, $$x=\sqrt{y}$$ We can express x as a function of y: $$x = y^{\frac{1}{2}}$$

Step 2: Substitute x in the second equation

Now, substitute x from Step 1 into the second equation: $$\left(y^{\frac{1}{2}}\right)^2 - 9y^2 = 9$$

Step 3: Solve the equation for y

Simplify the equation from Step 2: $$y - 9y^2 = 9$$ Rearrange the equation to form a quadratic equation: $$9y^2 - y - 9 = 0$$ To solve the quadratic equation, we can use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, a = 9, b = -1, and c = -9. Plug these values into the quadratic formula: $$y = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 9 \cdot (-9)}}{2 \cdot 9}$$

Step 4: Calculate the values of y

Calculate the values of y from the quadratic formula: $$y = \frac{1 \pm \sqrt{1 + 324}}{18}$$ $$y = \frac{1 \pm \sqrt{325}}{18}$$ There are two possible values for y: $$y_1 = \frac{1 + \sqrt{325}}{18}$$ $$y_2 = \frac{1 - \sqrt{325}}{18}$$

Step 5: Find the corresponding value(s) of x

Now, substitute these values of y into the equation derived in Step 1 to find the corresponding values of x: For \(y_1\): $$x_1 = \sqrt{y_1} = \sqrt{\frac{1 + \sqrt{325}}{18}}$$ For \(y_2\): $$x_2 = \sqrt{y_2} = \sqrt{\frac{1 - \sqrt{325}}{18}}$$ #Conclusion# The solutions of the given system of equations are: $$\left(x_1, y_1\right) = \left(\sqrt{\frac{1 + \sqrt{325}}{18}}, \frac{1 + \sqrt{325}}{18}\right)$$ $$\left(x_2, y_2\right) = \left(\sqrt{\frac{1 - \sqrt{325}}{18}}, \frac{1 - \sqrt{325}}{18}\right)$$

Key Concepts

Quadratic Equation

A quadratic equation is a type of algebraic equation where the highest power of the variable is squared. It takes the general form: \[ ax^2 + bx + c = 0 \]
In this equation, the letters \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The term \(ax^2\) represents the quadratic term, \(bx\) is the linear term, and \(c\) is the constant term.

• Quadratic equations can have either two distinct real roots, one real double root, or no real roots (two complex roots).
• The most common methods for solving these equations include: factoring, completing the square, or using the quadratic formula.

In our exercise, we encounter a quadratic equation after substituting in the system of equations. We rearrange the terms to form \(9y^2 - y - 9 = 0\) and use the quadratic formula to find the values of \(y\).

Substitution Method

The substitution method is one of the techniques used to solve a system of equations, especially when the system includes nonlinear equations like quadratic or higher. This method involves expressing one variable in terms of another from one equation and substituting it into the second equation.

• It is efficient for systems where it is easy to solve one of the equations for one variable.
• This method can convert a complicated system or nonlinear equation into a simpler one, often a single equation that can be solved easier.

In our scenario, we start by expressing \(x\) as \(\sqrt{y}\) from the first equation. Then substitute \(x = y^{1/2}\) into the second equation, which transforms our equations into a format that we can subsequently solve as a quadratic.

Solving Systems

Solving systems of equations involves finding all variable values that satisfy each equation in the system simultaneously. Systems can either be linear or non-linear.

• Linear systems can be tackled using methods such as substitution, elimination, or matrices.
• Non-linear systems generally require more complex approaches or iterative methods.

The primary goal is finding the point(s) of intersection, represented by the variable values. In our exercise, we are dealing with a non-linear system due to the presence of a square root equation and a quadratic equation. By using substitution, we simplified the problem to solving a single quadratic equation.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. It provides a structured way of forming and solving equations and expressions.

• Algebraic techniques are foundational as they allow for the solving of everyday mathematical problems and more complex problems in fields like engineering and physics.
• Core ideas in algebra include understanding expressions, equations, and the relationships between variables.

In our example, we see key algebraic processes: arranging equations, substituting expressions, and applying formulas (like the quadratic formula). These steps are typical operations in algebra to derive solutions systematically.