Problem 38 - College Algebra

Question

Solve each system by the method of your choice. $$\begin{aligned} &x^{2}-y^{2}-4 x+6 y-4=0\\\ &x^{2}+y^{2}-4 x-6 y+12=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the given system of equations are the points (1, -3) and (1, 5).

Step 1 - Reorganize the Equations

The first step is to rewrite the two given equations into a friendlier version. One of the best ways to approach this is to isolate \(x^{2}\) and \(y^{2}\) on one side of the equation and the rest of the terms on the other. Doing so, the equations become: \[x^{2}-y^{2}=4x-6y+4\] \[x^{2}+y^{2}=4x+6y-12\]

Step 2 - Use the Method of Substitution

One equation has \(x^{2}-y^{2}\) and the other \(x^{2}+y^{2}\), it becomes clear that adding these equations would eliminate \(y^{2}\) terms and by subtracting the second from first, we get rid of \(x^{2}\) terms. Let's first add the two re-organized equations: \[x^{2}-y^{2}+(x^{2}+y^{2})=4x-6y+4+(4x+6y-12)\] Simplifying this gives us: \[2x^{2}=8x-8\] Solving for \(x\), we get \(x=1\).

Step 3 - Substitute into one of the Equations to Solve for y

Substitute \(x=1\) into one of the equations, say the second one, to solve for \(y\). \[1+y^{2}=4+6y-12\] Solving this quadratic equation, we get \(y=-3,5\).

Step 4 - List all the Solutions

Finally, put \(x\) and \(y\)'s values together to give the coordinates of the intersection points. Therefore, the solutions to the system are the points with coordinates \((1,-3)\) and \((1,5)\).

Key Concepts

Substitution Method

The substitution method is a cornerstone algebraic technique employed for solving systems of equations. As the name implies, it involves substituting one equation into another. The main goal is to express one variable in terms of another, thereby reducing the complexity of the system. For instance, if you have a set of equations with variables, say, x and y, you first solve one equation for x in terms of y or vice versa. Then, you 'substitute' this expression into the other equation.

This method can significantly simplify problems, especially when dealing with linear and non-linear systems, such as the system of quadratic equations in the original exercise. When we isolated the terms containing x^2 and y^2 and then added the equations, we used substitution to make the other variable disappear, making it possible to solve for x without y interfering. Once x has been found, it can be substituted back into any of the original equations to find y. This method is highly effective but requires careful algebraic manipulation to ensure accuracy.

System of Quadratic Equations

A system of quadratic equations consists of multiple equations where the variables are raised to the second power or have products of different variables. Unlike linear systems, these equations graphically represent parabolas or other non-linear shapes. In the given exercise, both equations are non-linear and have terms in x^2 and y^2. Such systems often have more than one solution, corresponding to the intersection points of the curves represented by each equation.

To solve these systems sturdily, algebraic methods such as substitution, elimination, and graphing are used. In some cases, factoring or completing the square might be helpful as well. Systems of quadratic equations may have zero, one, or multiple solutions, and determining the number of solutions involves understanding the behavior of parabolas—they can intersect at no points, at one point (if they are tangent), or at two distinct points. By reorganizing and combining the equations algebraically, we can identify the solutions that satisfy both original equations. With the judicious use of algebraic techniques, even complex systems can be unraveled to find their solutions.

Algebraic Methods

Algebraic methods encompass a variety of techniques used to solve equations involving variables. These methods include but are not limited to substitution, elimination, factoring, and using the quadratic formula. The choice of method often depends on the form of the equation and the specific details of the problem at hand.

In our example problem, we're primarily focusing on the substitution and elimination methods. These methods work hand-in-hand, especially in systems where equations need to be manipulated so that one variable can be eliminated. By rearranging the initial equations to isolate terms, it becomes easier to see opportunities for substitution or elimination. Algebraic methods are powerful tools in mathematics that allow us to find precise solutions to systems of equations, no matter how challenging they may initially appear. As such, strengthening one's algebraic skills is invaluable for anyone aspiring to delve into more complex mathematical topics.