Solving Systems Of Equations Algebraically Step-by-Step

In the realm of mathematics, solving systems of equations is a fundamental skill with applications across various fields. A system of equations is a set of two or more equations with the same variables. The solution set of a system of equations is the set of all values that satisfy all equations simultaneously. There are several methods to solve systems of equations, including graphical, substitution, and elimination methods. However, in many cases, an algebraic approach provides the most precise and efficient means of finding the solution set. This article will delve into the initial steps involved in determining the solution set of a system of equations algebraically, using the given example of the equations $y = x^2 - 6x + 12$ and $y = 2x - 4$.

Step 1: Setting the Equations Equal to Each Other

The first critical step in solving this system of equations algebraically is to recognize that since both equations are expressed in terms of y, we can set their right-hand sides equal to each other. This is because, at the point(s) where the two equations intersect (i.e., the solution set), the y-values must be the same. By equating the expressions for y, we eliminate one variable (y) and obtain a single equation in terms of x. This allows us to focus on solving for x first, which will then pave the way for finding the corresponding y-values.

In our example, we have the equations:

• $y = x^2 - 6x + 12$
• $y = 2x - 4$

By setting the right-hand sides equal to each other, we get:

$x^2 - 6x + 12 = 2x - 4$

This resulting equation is a quadratic equation, which is an equation of the form $ax^2 + bx + c = 0$, where a, b, and c are constants. Quadratic equations can have up to two real solutions, which correspond to the x-coordinates of the points where the parabola represented by $y = x^2 - 6x + 12$ intersects the line represented by $y = 2x - 4$. The significance of this step cannot be overstated, as it transforms the problem from a system of two equations into a single equation that we can solve using standard algebraic techniques.

Understanding the Underlying Principle

The reason this step works lies in the fundamental concept of equality. If two expressions are equal to the same quantity, then they must be equal to each other. In this case, both $x^2 - 6x + 12$ and $2x - 4$ are equal to y at the points of intersection. Therefore, they must be equal to each other at those points. This principle is a cornerstone of algebraic manipulation and is used extensively in solving various types of equations.

By equating the expressions, we are essentially finding the x-values where the two graphs intersect. These x-values are the solutions to the equation $x^2 - 6x + 12 = 2x - 4$. Once we find these x-values, we can substitute them back into either of the original equations to find the corresponding y-values, thus completing the solution set.

Potential Pitfalls and Considerations

While setting the equations equal to each other is a straightforward step, it's important to be mindful of potential pitfalls. For instance, if the equations were more complex, involving radicals or fractions, extra care would be needed to avoid introducing extraneous solutions. Extraneous solutions are solutions that satisfy the transformed equation but not the original system of equations. These can arise due to the algebraic manipulations performed, such as squaring both sides of an equation.

In our specific case, since we are dealing with a quadratic equation, we need to be prepared for the possibility of having two, one, or no real solutions. This depends on the discriminant of the quadratic equation, which we will discuss in the next step.

Step 2: Rearranging the Equation into Standard Quadratic Form

Having successfully equated the two expressions for y, the next crucial step is to rearrange the resulting equation into the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. This standard form is essential because it allows us to readily apply various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. By transforming the equation into this format, we set the stage for a systematic approach to finding the solutions for x.

Starting from the equation we obtained in Step 1:

$x^2 - 6x + 12 = 2x - 4$

We need to manipulate this equation to get all terms on one side, leaving zero on the other side. To do this, we subtract $2x$ and add $4$ to both sides of the equation:

$x^2 - 6x + 12 - 2x + 4 = 2x - 4 - 2x + 4$

Simplifying the equation, we combine like terms:

$x^2 - 8x + 16 = 0$

Now, the equation is in the standard quadratic form, with $a = 1$, $b = -8$, and $c = 16$. This form provides a clear structure that makes it easier to identify the coefficients and constants needed for various solution methods.

Why Standard Form Matters

The standard quadratic form is not just a matter of convention; it serves a practical purpose. It organizes the terms in a way that highlights the key parameters of the quadratic equation. The coefficient a determines the parabola's concavity (whether it opens upwards or downwards) and its width. The coefficient b influences the parabola's axis of symmetry and horizontal position. The constant c represents the y-intercept of the parabola.

Furthermore, the standard form is essential for applying the quadratic formula, which is a general solution for any quadratic equation: $x = (-b ± √(b^2 - 4ac)) / (2a)$. This formula directly uses the coefficients a, b, and c to calculate the roots (solutions) of the equation. Similarly, when completing the square, the standard form helps in identifying the term needed to create a perfect square trinomial.

Factoring as a Solution Method

In some cases, the quadratic equation in standard form can be solved by factoring. Factoring involves expressing the quadratic expression as a product of two linear expressions. If we can factor the equation, we can then set each factor equal to zero and solve for x. In our example, the equation $x^2 - 8x + 16 = 0$ can be factored as:

$(x - 4)(x - 4) = 0$

This means that the equation has a repeated root, $x = 4$. Factoring is often the quickest method for solving quadratic equations when it is applicable, but it may not always be straightforward, especially when the coefficients are large or the roots are irrational.

The Discriminant and the Nature of Solutions

Before proceeding to solve for x, it's beneficial to consider the discriminant of the quadratic equation, which is given by the expression $b^2 - 4ac$. The discriminant provides information about the nature of the solutions. If the discriminant is positive, the equation has two distinct real roots. If it is zero, the equation has one repeated real root (as in our case). If it is negative, the equation has no real roots, but it has two complex roots.

In our example, the discriminant is:

$(-8)^2 - 4(1)(16) = 64 - 64 = 0$

This confirms that our equation has one repeated real root, which we found to be $x = 4$ by factoring. Understanding the discriminant can save time and effort by indicating the type of solutions to expect.

Subsequent Steps and the Complete Solution

Having completed the first two steps, we have successfully transformed the system of equations into a standard quadratic equation and determined the x-value(s) that satisfy it. The subsequent steps involve finding the corresponding y-value(s) and expressing the solution set in the appropriate format.

Step 3: Solving for x

As we saw in Step 2, the equation $x^2 - 8x + 16 = 0$ factors to $(x - 4)(x - 4) = 0$, which gives us the repeated solution $x = 4$. This means that the parabola and the line intersect at only one point.

Step 4: Solving for y

To find the y-value corresponding to $x = 4$, we can substitute this value into either of the original equations. Let's use the simpler equation, $y = 2x - 4$:

$y = 2(4) - 4 = 8 - 4 = 4$

So, the y-value is 4.

Step 5: Expressing the Solution Set

The solution set of the system of equations is the set of all ordered pairs (x, y) that satisfy both equations. In this case, we have only one solution, which is (4, 4). This means that the parabola and the line intersect at the point (4, 4).

Alternative Methods for Solving Quadratic Equations

While factoring worked well in this example, it's not always the most efficient method for solving quadratic equations. Two other common methods are completing the square and using the quadratic formula.

• Completing the Square: This method involves manipulating the quadratic equation to create a perfect square trinomial on one side. While it can be more involved than factoring, it is a powerful technique that can be used to solve any quadratic equation.

• Quadratic Formula: The quadratic formula, $x = (-b ± √(b^2 - 4ac)) / (2a)$, provides a direct way to find the solutions of a quadratic equation in standard form. It is particularly useful when factoring is difficult or impossible.

Conclusion

Solving systems of equations algebraically is a fundamental skill in mathematics. The first two steps, setting the equations equal to each other and rearranging the equation into standard quadratic form, are crucial for setting up the problem for solution. By understanding these steps and the underlying principles, you can effectively solve a wide range of systems of equations. The subsequent steps involve solving for x, solving for y, and expressing the solution set, leading to the complete solution of the system. Whether you choose to factor, complete the square, or use the quadratic formula, the key is to approach the problem systematically and methodically to arrive at the correct solution.