Solving Systems Of Equations: Quadratic And Linear

When you encounter a system of equations, especially one that involves a quadratic equation and a linear equation like the one presented:

$y = x^2 + 5x + 6$ $y = 3x + 6$

you're essentially looking for the points where these two distinct mathematical relationships intersect. Think of it like two paths crossing – we want to find the exact coordinates (x, y) where both paths are present simultaneously. The method to achieve this is elegantly straightforward: since both equations are set equal to 'y', we can set them equal to each other. This substitution is a powerful tool in algebra, allowing us to collapse two separate problems into one solvable equation. By equating the expressions for 'y', we transform the system into a single equation in terms of 'x', which we can then solve. This process is fundamental to understanding how different functions interact and where their common ground lies.

The core strategy for solving this particular system of equations lies in equating the two expressions for 'y'. Since both $y = x^2 + 5x + 6$ and $y = 3x + 6$ are given, we can directly set the right-hand sides of these equations equal to each other. This gives us: $x^2 + 5x + 6 = 3x + 6$. Now, our goal is to rearrange this equation into a standard quadratic form, which is typically $ax^2 + bx + c = 0$. To do this, we need to move all terms to one side of the equation, setting the other side to zero. Subtracting $3x$ from both sides, we get $x^2 + 2x + 6 = 6$. Next, subtracting $6$ from both sides yields $x^2 + 2x = 0$. This simplified quadratic equation is now ready for us to find its roots, which correspond to the x-coordinates of our intersection points. The ability to simplify and rearrange equations is a cornerstone of algebraic problem-solving, enabling us to tackle more complex scenarios by breaking them down into manageable steps. This initial algebraic manipulation is crucial for isolating the variable and preparing the equation for further analysis, whether through factoring, using the quadratic formula, or other solution methods. The elegance of this step is that it directly links the graphical intersection points to the algebraic solutions of the derived equation, providing a unified understanding of the problem.

Once we have the equation $x^2 + 2x = 0$, we need to find the values of 'x' that satisfy it. This specific quadratic equation is relatively easy to solve by factoring. We can see that both terms, $x^2$ and $2x$, share a common factor of 'x'. By factoring out 'x', we get $x(x + 2) = 0$. The Zero Product Property states that if the product of two or more factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero: $x = 0$ or $x + 2 = 0$. Solving the second equation, $x + 2 = 0$, by subtracting $2$ from both sides gives us $x = -2$. So, we have found two possible x-values for the intersection points: $x = 0$ and $x = -2$. These are the first part of our solutions. Finding these roots is a critical step, as they represent the horizontal positions where the parabola and the line meet. Factoring is often the most efficient method when applicable, but if factoring proves difficult, the quadratic formula ($x = [-b \pm \sqrt{b^2 - 4ac}] / 2a$) is a universal tool that can solve any quadratic equation. The importance of these x-values cannot be overstated; they are the direct keys to unlocking the full coordinates of the intersection points, bridging the gap between the abstract algebraic manipulation and the concrete graphical representation of the solution.

Having found the x-coordinates ($x = 0$ and $x = -2$), the next logical step is to find the corresponding y-coordinates for each of these x-values. To do this, we can substitute each x-value back into either of the original equations. The linear equation, $y = 3x + 6$, is usually simpler to work with, making calculations quicker and less prone to error. Let's start with $x = 0$. Substituting this into the linear equation gives us $y = 3(0) + 6$, which simplifies to $y = 0 + 6$, so $y = 6$. This means one of our intersection points is $(0, 6)$. Now, let's take the other x-value, $x = -2$. Substituting this into the linear equation gives us $y = 3(-2) + 6$, which simplifies to $y = -6 + 6$, so $y = 0$. This gives us our second intersection point: $(-2, 0)$. It's always a good practice to verify these points by substituting them into the other original equation (the quadratic one in this case) to ensure consistency. For $(0, 6)$: $6 = (0)^2 + 5(0) + 6 ightarrow 6 = 0 + 0 + 6 ightarrow 6 = 6$. This checks out. For $(-2, 0)$: $0 = (-2)^2 + 5(-2) + 6 ightarrow 0 = 4 - 10 + 6 ightarrow 0 = 0$. This also checks out. This verification step is crucial for confirming the accuracy of our solutions and reinforces the understanding that these points lie on both the parabola and the line. Finding the y-coordinates completes the solution set, providing the precise locations where the two graphs intersect.

Looking at the options provided, we have identified our solution points as $(0, 6)$ and $(-2, 0)$. Let's compare this with the given choices:

A. ( $-3,0$ ) and ( 0,6 ) B. $(0,6)$ and $(-2,0)$ C. $(-3,0)$ and $(-2,0)

Our calculated solutions perfectly match option B. This confirms that the points $(0,6)$ and $(-2,0)$ are indeed the solutions to the given system of equations. The process of solving systems of equations, whether graphical or algebraic, is fundamental in mathematics and has wide-ranging applications. It's used in fields like economics to find market equilibrium, in physics to analyze projectile motion and forces, and in engineering to model complex systems. Understanding how to find these intersection points provides insight into the behavior and relationships between different mathematical models. The satisfaction of arriving at the correct solution, especially when it aligns with one of the provided options, is a testament to the power and logic of algebraic manipulation. This methodical approach ensures that we leave no stone unturned in our quest for mathematical accuracy.

Why These Points Matter

The solutions we found, $(0,6)$ and $(-2,0)$, represent the points of intersection between the graph of the quadratic equation $y = x^2 + 5x + 6$ (a parabola) and the graph of the linear equation $y = 3x + 6$ (a straight line). Visualizing this can be incredibly helpful. The parabola opens upwards, and the line has a positive slope. The fact that we found two distinct points of intersection means the line cuts through the parabola at two different locations. The point $(0,6)$ is where the line and parabola both cross the y-axis. The point $(-2,0)$ is where the line and the parabola both cross the x-axis. If we had found only one solution, it would mean the line is tangent to the parabola (just touching it at one point). If we had found no real solutions, it would mean the line and the parabola do not intersect at all. The number of solutions depends on the specific coefficients in the equations, and the algebraic methods we used (substitution and solving the resulting quadratic equation) directly reveal this number and the coordinates of the intersection points. Understanding these intersections is not just an academic exercise; it's key to modeling real-world phenomena where two different trends or constraints meet. For instance, in economics, it could represent the price point where supply equals demand. In physics, it might be the point where a thrown object hits the ground after being launched along a certain trajectory. The graphical interpretation solidifies the algebraic results and provides a more intuitive grasp of the problem's context. The precision of these points is vital for accurate predictions and analyses in any applied field.

Alternative Solution Methods

While substitution and factoring (or the quadratic formula) are standard for solving systems like this, it's worth noting other approaches. Graphical methods can provide an approximate solution. If you were to plot both $y = x^2 + 5x + 6$ and $y = 3x + 6$ on the same coordinate plane, you could visually identify the points where the graphs intersect. This method is excellent for conceptual understanding and for getting a rough idea of the solution, but it might not yield exact coordinates unless the intersection points have simple integer or fractional values and the graph is drawn very precisely. For more complex equations or when exactness is paramount, algebraic methods are superior. Another algebraic technique, though less direct for this specific problem, is elimination. However, elimination is typically more useful when dealing with systems of linear equations, or when the equations can be manipulated to eliminate a variable directly. In this case, since both equations are already solved for 'y', substitution is the most natural and efficient method. The choice of method often depends on the form of the equations and the desired level of precision. For instance, if the question asked for an approximate solution, graphing would be a valid strategy. But given the multiple-choice format with precise coordinate pairs, an exact algebraic solution is necessary. Always consider the nature of the equations and the required output when selecting your problem-solving strategy. The robustness of mathematics lies in its diverse toolkit, allowing us to select the most appropriate instrument for each task.

Conclusion

Solving systems of equations where one is quadratic and the other is linear involves a systematic approach. We begin by using substitution to set the expressions for 'y' equal to each other, transforming the system into a single quadratic equation. This equation, $x^2 + 2x = 0$, was then solved by factoring, yielding the x-coordinates $x = 0$ and $x = -2$. By substituting these x-values back into the simpler linear equation, we found the corresponding y-coordinates, resulting in the intersection points $(0, 6)$ and $(-2, 0)$. These points represent the exact locations where the parabola and the line meet on a graph. The process highlights the interplay between algebra and geometry, where algebraic solutions correspond to graphical intersection points. Mastery of these techniques is essential for tackling more advanced mathematical concepts and real-world applications.

For further exploration into the world of algebra and systems of equations, you can visit the Khan Academy website, which offers extensive resources and practice problems. Additionally, the Math is Fun website provides clear explanations and interactive tools related to algebraic concepts.