Solving System Of Equations F(x) = 3x + 6 And G(x) = X² + 4x + 1
In the realm of mathematics, solving systems of equations is a fundamental skill with applications across various fields. This article delves into the process of finding solutions for a system of equations, specifically focusing on the equations f(x) = 3x + 6 and g(x) = x² + 4x + 1. We will explore different methods to solve this system, providing a step-by-step guide to help you understand the underlying concepts and arrive at accurate solutions. Furthermore, we'll address the importance of rounding solutions to the nearest tenth when necessary, ensuring practical applicability in real-world scenarios.
Understanding the System of Equations
Before diving into the solution methods, let's first grasp the essence of the system of equations we're dealing with. We have two equations:
- f(x) = 3x + 6
- g(x) = x² + 4x + 1
The first equation, f(x) = 3x + 6, represents a linear function. Linear functions, characterized by their straight-line graphs, have a constant rate of change. In this case, the slope of the line is 3, and the y-intercept is 6. This means that for every unit increase in x, the value of f(x) increases by 3, and the line crosses the y-axis at the point (0, 6).
The second equation, g(x) = x² + 4x + 1, represents a quadratic function. Quadratic functions are distinguished by their parabolic graphs, which are U-shaped curves. The general form of a quadratic function is ax² + bx + c, where a, b, and c are constants. In our equation, a = 1, b = 4, and c = 1. The solutions to a quadratic equation, also known as the roots or zeros, are the x-values where the parabola intersects the x-axis.
Solving a system of equations involves finding the x-values that satisfy both equations simultaneously. In other words, we're looking for the points where the graphs of the two functions intersect. These points of intersection represent the solutions to the system.
Methods for Solving the System of Equations
Several methods can be employed to solve a system of equations. In this section, we'll focus on two common approaches: the substitution method and the graphical method.
1. Substitution Method
The substitution method is an algebraic technique that involves solving one equation for one variable and substituting that expression into the other equation. This process eliminates one variable, resulting in a single equation that can be solved for the remaining variable. Let's apply this method to our system of equations.
Step 1: Solve one equation for one variable.
We can choose either equation to solve for either variable. In this case, let's solve the first equation, f(x) = 3x + 6, for x. Subtracting 6 from both sides, we get:
3x = f(x) - 6
Dividing both sides by 3, we have:
x = (f(x) - 6) / 3
Step 2: Substitute the expression into the other equation.
Now, we substitute this expression for x into the second equation, g(x) = x² + 4x + 1:
g(x) = ((f(x) - 6) / 3)² + 4((f(x) - 6) / 3) + 1
Step 3: Simplify and solve the equation.
This equation now contains only one variable, f(x). Let's simplify and solve for f(x). First, let's replace g(x) with f(x) since we are looking for the intersection points where f(x) = g(x):
f(x) = ((f(x) - 6) / 3)² + 4((f(x) - 6) / 3) + 1
Let's use 'y' to represent f(x) to simplify the equation:
y = ((y - 6) / 3)² + 4((y - 6) / 3) + 1
Multiply both sides by 9 to eliminate fractions:
9y = (y - 6)² + 12(y - 6) + 9
Expand and simplify:
9y = y² - 12y + 36 + 12y - 72 + 9
9y = y² - 27
Rearrange the equation into a quadratic form:
0 = y² - 9y - 27
Now, we can use the quadratic formula to solve for y:
y = (-b ± √(b² - 4ac)) / 2a
Where a = 1, b = -9, and c = -27.
y = (9 ± √((-9)² - 4(1)(-27))) / 2(1)
y = (9 ± √(81 + 108)) / 2
y = (9 ± √189) / 2
y ≈ (9 ± 13.75) / 2
So, we have two possible values for y:
y₁ ≈ (9 + 13.75) / 2 ≈ 11.375
y₂ ≈ (9 - 13.75) / 2 ≈ -2.375
Step 4: Substitute the values back to find the other variable.
Now that we have the values for y (which is f(x)), we can substitute them back into either of the original equations to find the corresponding x values. Let's use the simpler equation, f(x) = 3x + 6.
For y₁ ≈ 11.375:
11.375 = 3x + 6
5.375 = 3x
x₁ ≈ 1.79
For y₂ ≈ -2.375:
-2.375 = 3x + 6
-8.375 = 3x
x₂ ≈ -2.79
Therefore, the solutions are approximately (1.79, 11.375) and (-2.79, -2.375).
2. Graphical Method
The graphical method involves plotting the graphs of both equations on the same coordinate plane. The points where the graphs intersect represent the solutions to the system of equations. This method provides a visual representation of the solutions and can be particularly useful for understanding the nature of the system.
Step 1: Plot the graphs of the equations.
To plot the graph of f(x) = 3x + 6, we can identify two points on the line. For example, when x = 0, f(x) = 6, and when x = -2, f(x) = 0. Plotting these points (0, 6) and (-2, 0) and drawing a line through them gives us the graph of f(x).
To plot the graph of g(x) = x² + 4x + 1, we can identify the vertex and a few other points. The x-coordinate of the vertex is given by -b/2a, where a = 1 and b = 4. So, the x-coordinate of the vertex is -4/2(1) = -2. The y-coordinate of the vertex is g(-2) = (-2)² + 4(-2) + 1 = -3. Therefore, the vertex is at (-2, -3). We can also find the x-intercepts by setting g(x) = 0 and solving for x using the quadratic formula. These points, along with a few other points, allow us to sketch the parabola.
Step 2: Identify the points of intersection.
Once we have plotted the graphs of both equations, we can visually identify the points where they intersect. These points represent the solutions to the system of equations. From the graph, we can see that the two curves intersect at approximately (1.8, 11.4) and (-2.8, -2.4).
Step 3: Verify the solutions.
To ensure the accuracy of our solutions, we can substitute the x and y values back into the original equations. If both equations are satisfied, then the point is indeed a solution.
Rounding to the Nearest Tenth
In many practical applications, it's necessary to round solutions to a specific decimal place. When rounding to the nearest tenth, we look at the digit in the hundredths place. If it's 5 or greater, we round the tenths digit up. If it's less than 5, we leave the tenths digit as it is.
Applying this to our solutions obtained from the substitution method (1.79, 11.375) and (-2.79, -2.375), we round to the nearest tenth as follows:
- (1.79, 11.375) rounds to (1.8, 11.4)
- (-2.79, -2.375) rounds to (-2.8, -2.4)
These rounded solutions align with the approximate solutions obtained from the graphical method, further validating our results.
Conclusion
Solving systems of equations is a fundamental mathematical skill with diverse applications. In this article, we explored the process of finding solutions for the system of equations f(x) = 3x + 6 and g(x) = x² + 4x + 1. We employed both the substitution method and the graphical method, demonstrating the algebraic and visual approaches to solving such systems. Additionally, we emphasized the importance of rounding solutions to the nearest tenth when necessary, ensuring practical relevance in real-world contexts.
By mastering the techniques discussed in this article, you'll gain a solid foundation for solving systems of equations, a skill that will prove invaluable in various mathematical and scientific endeavors. Remember to practice these methods with different systems of equations to enhance your understanding and proficiency. With consistent effort, you'll become adept at finding solutions and interpreting their significance in diverse scenarios.
