Problem 20 - College Algebra Essentials

Question

In Exercises \(19-28,\) solve each system by the addition method. $$ \left\\{\begin{array}{l} 4 x^{2}-y^{2}=4 \\ 4 x^{2}+y^{2}=4 \end{array}\right. $$

Step-by-Step Solution

Answer

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

Step 1 - Combine the Equations

By the addition method, one can add both equations together. The equation thus formed will be: \(4x^2 - y^2 + 4x^2 + y^2 = 4 + 4\). See here, the \(y^2\) term in the equation cancels out, leaving us with the equation: \(8x^2 = 8\).

Step 2 - Solve for Variable x

After simplification of the equation from step 1, solving for x becomes straightforward. Simplify the equation by dividing both sides by 8 to isolate x: \(x^2 = 1\). Therefore, taking square roots gives the possible solutions as \(x = -1\) and \(x = 1\).

Step 3 - Substitute x Values for y and Solve

Substitute the x values back into the original equations to find the y: Substitute \(x = 1\) into either of the original equations, resulting in \(4 - y^2 = 4\), which yields \(y = 0\) as the root. Similarly, when \(x = -1\) is substituted, \(y\) equals \(0\) as well. Hence, our solutions are \((-1, 0)\) and \((1, 0)\).

Key Concepts

System of Equations

A system of equations consists of two or more equations that share the same set of variables. In this exercise, the system is comprised of two quadratic equations:

• \(4x^2 - y^2 = 4\)
• \(4x^2 + y^2 = 4\)

These equations are linked together by the variables \(x\) and \(y\). The goal is to find values of \(x\) and \(y\) that satisfy both equations simultaneously. Solving such systems often reveals points where the graphs of the equations intersect. This can represent solutions on a graph where the equations are plotted. In this system, the method of choice is the addition (or elimination) method, in which the equations are combined in a way that simplifies the process of finding solutions.

Quadratic Equations

Quadratic equations are polynomial equations of degree two and have the standard form \(ax^2 + bx + c = 0\). In the given system, both equations are quadratic in terms of \(x\) and \(y\). Specifically, they consist of terms like \(x^2\) and \(y^2\), but importantly they do not include linear terms involving \(x\) or \(y\) alone. Quadratic equations are particularly interesting because they:

• Can have zero, one, or two real roots.
• Graph as parabolas, which can open upward or downward depending on the sign of the coefficient multiplying the square term.

In our system, the parabolas described by the two quadratic equations intersect at specific points, representing the solutions found. These intersections are determined by finding the common solutions for \(x\) and \(y\) that satisfy both quadratic equations.

Solution Method

The addition method, also known as the elimination method, involves adding or subtracting equations to eliminate one variable, making it easier to solve for the other. Here's a breakdown of how it works in this problem:- **Initial Equations**: Begin with the two given equations: \(4x^2 - y^2 = 4\) \(4x^2 + y^2 = 4\)- **Adding Equations**: Add both equations together. The \(y^2\) terms cancel out, simplifying the system to a single equation: \(8x^2 = 8\).- **Solving for \(x\)**: Solve \(8x^2 = 8\) by dividing each side by 8, which simplifies to \(x^2 = 1\). Solving for \(x\) yields the solutions \(x = 1\) and \(x = -1\), by taking the square root of both sides.- **Substituting back for \(y\)**: For each value of \(x\), substitute back into one of the original equations to solve for \(y\). In this case, both values of \(x\) give \(y = 0\) upon substitution, providing the solution points \((-1, 0)\) and \((1, 0)\). The addition method is quite powerful for systems of equations, especially when the goal is to eliminate a variable quickly and efficiently. It is widely used for solving linear systems as well as simplifying certain types of nonlinear systems.