Solving Systems Of Equations A Comprehensive Guide

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In the realm of mathematics, solving systems of equations is a fundamental skill with applications spanning various fields, from engineering to economics. This article delves into the intricacies of solving a specific system of equations, providing a step-by-step approach and highlighting key concepts along the way. We will explore the given system:

y = x^2 - 2x - 19
y + 4x = 5

and determine the points that represent its solution set. This exploration will not only provide the solution to this particular problem but also equip you with the tools to tackle similar challenges.

Understanding the System of Equations

At its core, a system of equations is a collection of two or more equations that share the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. In our case, we have a system consisting of two equations:

  1. A quadratic equation: y = x^2 - 2x - 19
  2. A linear equation: y + 4x = 5

The first equation represents a parabola, while the second equation represents a straight line. The solutions to this system correspond to the points where the parabola and the line intersect on a graph. These points of intersection represent the (x, y) coordinates that satisfy both equations.

To effectively solve this system, we need to employ algebraic techniques that allow us to manipulate the equations and isolate the variables. The most common methods for solving systems of equations include substitution, elimination, and graphical methods. In this instance, we will utilize the substitution method, as it is particularly well-suited for systems where one equation is already solved for one variable in terms of the other.

The Substitution Method: A Step-by-Step Approach

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This process reduces the system to a single equation with a single variable, which can then be solved using standard algebraic techniques. Once the value of one variable is found, it can be substituted back into either of the original equations to find the value of the other variable.

In our system, the first equation, y = x^2 - 2x - 19, is already solved for y in terms of x. This makes the substitution method a natural choice. We can substitute the expression x^2 - 2x - 19 for y in the second equation, y + 4x = 5. This substitution yields:

(x^2 - 2x - 19) + 4x = 5

Now we have a single equation with only one variable, x. This equation can be simplified and solved for x.

Solving the Quadratic Equation

After substituting, we obtain the equation:

x^2 - 2x - 19 + 4x = 5

Combining like terms, we get:

x^2 + 2x - 19 = 5

To solve this quadratic equation, we need to rearrange it into the standard form of a quadratic equation, which is ax^2 + bx + c = 0. Subtracting 5 from both sides, we get:

x^2 + 2x - 24 = 0

Now we have a quadratic equation in standard form. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, the equation can be readily factored.

Factoring the Quadratic Equation

Factoring involves expressing the quadratic equation as a product of two binomials. To factor x^2 + 2x - 24 = 0, we need to find two numbers that multiply to -24 and add up to 2. These numbers are 6 and -4. Therefore, we can factor the equation as:

(x + 6)(x - 4) = 0

For the product of two factors to be zero, at least one of the factors must be zero. This leads to two possible solutions for x:

  1. x + 6 = 0, which gives x = -6
  2. x - 4 = 0, which gives x = 4

So, we have found two possible values for x: -6 and 4. These values correspond to the x-coordinates of the points where the parabola and the line intersect.

Finding the Corresponding y-values

Now that we have the x-values, we need to find the corresponding y-values. We can do this by substituting each x-value back into either of the original equations. The linear equation y + 4x = 5 is generally easier to work with. Let's substitute each x-value into this equation:

  1. For x = -6:

    y + 4(-6) = 5 y - 24 = 5 y = 29

    This gives us the solution point (-6, 29), which was already provided.

  2. For x = 4:

    y + 4(4) = 5 y + 16 = 5 y = -11

    This gives us the solution point (4, -11).

Therefore, the two points representing the solution set of the system of equations are (-6, 29) and (4, -11).

Verifying the Solutions

It's always a good practice to verify the solutions by substituting them back into both original equations to ensure they satisfy both. Let's verify our solutions:

  1. For the point (-6, 29):

    • Equation 1: y = x^2 - 2x - 19

      29 = (-6)^2 - 2(-6) - 19 29 = 36 + 12 - 19 29 = 29 (This equation is satisfied)

    • Equation 2: y + 4x = 5

      29 + 4(-6) = 5 29 - 24 = 5 5 = 5 (This equation is satisfied)

  2. For the point (4, -11):

    • Equation 1: y = x^2 - 2x - 19

      -11 = (4)^2 - 2(4) - 19 -11 = 16 - 8 - 19 -11 = -11 (This equation is satisfied)

    • Equation 2: y + 4x = 5

      -11 + 4(4) = 5 -11 + 16 = 5 5 = 5 (This equation is satisfied)

Both points satisfy both equations, confirming that they are indeed the solutions to the system of equations.

Graphical Interpretation of the Solutions

The solutions to a system of equations can also be visualized graphically. The graph of the quadratic equation y = x^2 - 2x - 19 is a parabola, and the graph of the linear equation y + 4x = 5 is a straight line. The points of intersection of these two graphs represent the solutions to the system of equations.

If you were to plot these two equations on a coordinate plane, you would observe that the parabola and the line intersect at two points: (-6, 29) and (4, -11). This graphical representation provides a visual confirmation of our algebraic solutions.

Real-World Applications of Systems of Equations

Systems of equations are not just abstract mathematical concepts; they have numerous real-world applications. They are used to model and solve problems in various fields, including:

  • Engineering: Systems of equations are used to analyze circuits, design structures, and model fluid flow.
  • Economics: They are used to model supply and demand, analyze market equilibrium, and forecast economic trends.
  • Physics: Systems of equations are used to describe the motion of objects, analyze forces, and model energy transfer.
  • Computer Science: They are used in optimization algorithms, computer graphics, and machine learning.

For instance, consider a business scenario where a company wants to determine the optimal pricing strategy for two products. The demand for each product may depend on the price of both products, leading to a system of equations that can be solved to find the prices that maximize profit. Another example is in circuit analysis, where Kirchhoff's laws lead to a system of equations that can be solved to determine the currents and voltages in different parts of the circuit.

Conclusion: Mastering Systems of Equations

In this article, we have explored the process of solving a system of equations consisting of a quadratic equation and a linear equation. We employed the substitution method to reduce the system to a single quadratic equation, which we then solved by factoring. We found the two solution points, (-6, 29) and (4, -11), and verified their correctness by substituting them back into the original equations. We also discussed the graphical interpretation of the solutions and highlighted the numerous real-world applications of systems of equations.

Mastering the techniques for solving systems of equations is an essential skill in mathematics and beyond. The ability to solve these systems empowers you to tackle a wide range of problems in science, engineering, economics, and other fields. By understanding the underlying concepts and practicing different solution methods, you can confidently approach and solve even the most challenging systems of equations.

The problem presented a system of equations:

y = x^2 - 2x - 19
y + 4x = 5

We were given one solution (-6, 29) and asked to find the other. Through the steps outlined above, we successfully identified the second solution as (4, -11).