Solving System Of Equations Algebraically Step-by-Step Guide

Solving systems of equations algebraically involves finding the points where the graphs of the equations intersect. This means finding the values of the variables that satisfy all equations in the system simultaneously. When dealing with a system of equations where one equation is quadratic and the other is linear, a common approach involves substitution and solving the resulting quadratic equation. In this article, we will delve into the steps required to solve the system of equations: $y = -x^2 - 4x - 3$ and $y = 2x + 5$ algebraically, with a focus on detailing the next step after equating the two equations.

Setting the Stage: Equating the Equations

The first crucial step in solving this system algebraically is recognizing that at the points of intersection, the y-values of both equations must be equal. This allows us to set the two expressions for y equal to each other, forming a single equation in terms of x. This foundational step transforms the system of two equations into a single equation that we can manipulate and solve. Therefore, as stated, we begin by setting the two equations equal to each other:

$-x^2 - 4x - 3 = 2x + 5$

This equation represents the x-values where the parabola defined by $y = -x^2 - 4x - 3$ and the line defined by $y = 2x + 5$ intersect. Solving this equation will give us the x-coordinates of these intersection points. Understanding this initial step is essential for navigating the rest of the solution process. Now, let's move on to the crucial next step.

The Next Step: Transforming to Standard Quadratic Form

After setting the two equations equal, the next logical and essential step is to rearrange the resulting equation into the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. This form is crucial because it allows us to apply various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. To achieve this standard form, we need to manipulate the equation by moving all terms to one side, leaving zero on the other side. This involves combining like terms and simplifying the equation.

In our case, we have the equation $-x^2 - 4x - 3 = 2x + 5$. To get it into standard form, we want to move all terms to the left side of the equation. We can do this by subtracting $2x$ and $5$ from both sides:

$-x^2 - 4x - 3 - 2x - 5 = 2x + 5 - 2x - 5$

This simplifies to:

$-x^2 - 6x - 8 = 0$

Now, to make the leading coefficient positive (which is often preferred for easier manipulation), we can multiply both sides of the equation by -1:

$(-1)(-x^2 - 6x - 8) = (-1)(0)$

This gives us:

$x^2 + 6x + 8 = 0$

This is the standard form of the quadratic equation, and it sets the stage for the subsequent steps in solving for x. Recognizing the importance of the standard form is key to effectively solving quadratic equations. From here, we can choose the most appropriate method to find the values of x that satisfy the equation. The transformation to standard form is not just a matter of algebraic manipulation; it's a strategic step that unlocks the potential to apply powerful solution techniques.

Solving the Quadratic Equation: Factoring

Now that we have the quadratic equation in standard form, $x^2 + 6x + 8 = 0$, we can proceed to solve for x. One of the most efficient methods for solving quadratic equations is factoring, if the equation is factorable. Factoring involves expressing the quadratic expression as a product of two binomials. To factor the quadratic expression $x^2 + 6x + 8$, we need to find two numbers that multiply to 8 (the constant term) and add up to 6 (the coefficient of the x term). These numbers are 4 and 2, since 4 * 2 = 8 and 4 + 2 = 6. Therefore, we can factor the quadratic expression as follows:

$x^2 + 6x + 8 = (x + 4)(x + 2)$

Now our equation becomes:

$(x + 4)(x + 2) = 0$

According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. This means that either $(x + 4) = 0$ or $(x + 2) = 0$. Solving each of these linear equations gives us the x-values that satisfy the quadratic equation:

For $x + 4 = 0$, we subtract 4 from both sides to get $x = -4$.

For $x + 2 = 0$, we subtract 2 from both sides to get $x = -2$.

Thus, we have found two solutions for x: $x = -4$ and $x = -2$. These are the x-coordinates of the points where the parabola and the line intersect. Factoring is a powerful technique because it directly leads to the solutions, providing a clear and concise path to the x-values. However, it's important to remember that not all quadratic equations are factorable, and in those cases, other methods like the quadratic formula or completing the square are necessary.

Finding the Corresponding y-Values

Having determined the x-values where the parabola and the line intersect, the next crucial step is to find the corresponding y-values. These y-values, along with the x-values we found, will give us the complete coordinates of the intersection points. To find the y-values, we can substitute each x-value back into either of the original equations. It's often easier to use the linear equation, $y = 2x + 5$, as it involves simpler calculations.

Let's start with $x = -4$. Substituting this value into the linear equation gives us:

$y = 2(-4) + 5$

$y = -8 + 5$

$y = -3$

So, when $x = -4$, the corresponding y-value is -3. This gives us one intersection point at (-4, -3).

Next, let's substitute $x = -2$ into the linear equation:

$y = 2(-2) + 5$

$y = -4 + 5$

$y = 1$

Therefore, when $x = -2$, the corresponding y-value is 1. This gives us the second intersection point at (-2, 1).

By substituting the x-values back into one of the original equations, we have successfully found the y-values that complete the solution. This step is vital because it provides the full picture of where the two graphs intersect, giving us the complete solution to the system of equations. The coordinates (-4, -3) and (-2, 1) represent the points where the parabola and the line cross each other on the coordinate plane. This process of finding corresponding y-values is a fundamental aspect of solving systems of equations graphically and algebraically.

Solutions and Interpretation

We have now successfully navigated the algebraic process of solving the system of equations $y = -x^2 - 4x - 3$ and $y = 2x + 5$. By setting the equations equal, transforming to standard quadratic form, factoring the quadratic equation, and finding the corresponding y-values, we have arrived at the solutions. The solutions to the system are the points of intersection of the parabola and the line, which we found to be (-4, -3) and (-2, 1).

These points represent the ordered pairs (x, y) that satisfy both equations simultaneously. In other words, if we substitute these x and y values into both equations, the equations will hold true. This is the fundamental concept behind solving systems of equations – finding the values that make all equations in the system true at the same time.

Graphically, these solutions represent the points where the parabola and the line intersect on the coordinate plane. The parabola opens downwards due to the negative coefficient of the $x^2$ term, and the line has a positive slope. The two points of intersection indicate that the line crosses the parabola at two distinct locations. Understanding the graphical interpretation reinforces the algebraic solution and provides a visual representation of the relationship between the two equations.

In conclusion, solving systems of equations algebraically is a powerful tool for finding solutions that satisfy multiple equations. The steps involved, such as equating equations, transforming to standard form, solving for variables, and interpreting the results, are essential skills in mathematics and have applications in various fields, including physics, engineering, and economics. This step-by-step guide provides a clear understanding of the process and highlights the significance of each step in arriving at the final solutions.