Solve Systems By Substitution: A Step-by-Step Guide
Solving systems of equations by substitution is a fundamental skill in algebra, offering a powerful method to find the point where two lines intersect. This technique is particularly useful when one of the equations is already solved for one variable, making the substitution process straightforward. Today, we'll dive deep into how to efficiently solve systems using substitution, breaking down the process into easy-to-follow steps and exploring why this method is so effective. We'll tackle a specific example to illustrate the concepts, ensuring you feel confident in applying this algebraic tool. Our goal is to demystify substitution, transforming it from a daunting task into a clear and manageable strategy for finding solutions to simultaneous equations.
Understanding the Substitution Method
The substitution method is an algebraic technique used to solve systems of linear equations. Its core principle lies in replacing a variable in one equation with an expression derived from another equation. This process effectively reduces a system of two equations with two variables into a single equation with a single variable, which can then be solved directly. The beauty of substitution is its versatility; it works whether the equations are presented in standard form () or slope-intercept form (). The key requirement for this method to be most effective is that at least one equation must be easily isolatable for one of its variables. This means an equation where a variable has a coefficient of 1 or -1, or an equation that is already solved for a variable (like or ). When you can isolate a variable, you create a direct link between the two equations, allowing you to substitute its equivalent expression from one equation into the other.
Before diving into the steps, let's briefly touch on why substitution is so valuable. Unlike graphical methods, which can be imprecise due to drawing errors or the nature of irrational solutions, the substitution method provides exact algebraic solutions. It's also more systematic than elimination when equations aren't neatly aligned for variable cancellation. The process involves a logical flow: isolate, substitute, solve, and back-substitute. Each step builds upon the previous one, leading you systematically towards the unique solution (or solutions, if they exist) of the system. Mastering this method not only helps in solving these specific problems but also builds a stronger foundation for more complex algebraic manipulations and problem-solving scenarios you'll encounter later on.
Step-by-Step Guide to Substitution
Let's walk through the process of solving a system of equations using the substitution method. We'll use the example provided: $eginarray}{l} y=-6 x \ y=-8 x-14 ext { ullet } ext { ullet } ext { ullet } ext { Step 1 ext ullet } ext { ullet } ext { ullet } ext { In our example, both equations are already conveniently solved for } y ext {. This makes our first step incredibly easy! We can choose either equation to work with, as they both give us an expression for } y ext { that we can use. If one equation was, say, } 2x + y = 5 ext {, and the other was } y = 3x - 1 ext {, we would immediately choose the second equation because } y ext { is already isolated. If neither equation was isolated, we would look for the variable with the simplest coefficient (usually 1 or -1) in either equation and solve for it. For instance, if we had } x + 2y = 7 ext {, we could easily solve for } x ext { to get } x = 7 - 2y ext {. } ext { ullet } ext { ullet } ext { ullet } ext { Step 2 ext ullet } ext { ullet } ext { ullet } ext { Now, we take the expression for } y ext { from the first equation, which is } y = -6x ext {, and substitute it into the second equation wherever we see } y ext {. Our second equation is } y = -8x - 14 ext {. So, replacing } y ext { with } -6x ext{ gives us -6x = -8x - 14 ext . The goal here is to eliminate one variable from one of the equations, leaving you with an equation that only contains the other variable. This is the crucial 'substitution' step that gives the method its name. } ext { ullet } ext { ullet } ext { ullet } ext { Step 3 ext ullet } ext { ullet } ext { ullet } ext { We now have a single equation with only } x ext{ as the variable -6x = -8x - 14 ext . To solve for } x ext{, we need to gather all the } x ext{ terms on one side of the equation and the constants on the other. Let's add } 8x ext{ to both sides -6x + 8x = -8x + 8x - 14 ext . This simplifies to } 2x = -14 ext {. Finally, we divide both sides by 2 to isolate } x ext{ x = rac-14}{2} ext {, which means } x = -7 ext {. We have successfully found the value of one of our variables! } ext { ullet } ext { ullet } ext { ullet } ext { Step 4 ext ullet } ext { ullet } ext { ullet } ext { Now that we know } x = -7 ext{, we can substitute this value back into either of the original equations to find the value of } y ext{. Using the first equation, } y = -6x ext{, is usually the easiest choice when a variable is already isolated. Substituting } -7 ext{ for } x ext{ gives us y = -6(-7) ext. Performing the multiplication, we get } y = 42 ext{. If we had used the second equation, } y = -8x - 14 ext{, we would get } y = -8(-7) - 14 ext{, which simplifies to } y = 56 - 14 ext{, resulting in } y = 42 ext{ as well. The fact that both equations yield the same } y ext{ value confirms our solution. } ext { ullet } ext { ullet } ext { ullet } ext { Step 5 ext ullet } ext { ullet } ext { ullet } ext { The solution to a system of linear equations is the point where the lines represented by the equations intersect. This point is expressed as an ordered pair } (x, y) ext{. Based on our calculations, we found } x = -7 ext{ and } y = 42 ext{. Therefore, the solution to the system is } (-7, 42) ext{. } ext { ullet } ext { ullet } ext { ullet } ext { Step 6 ext ullet } ext { ullet } ext { ullet } ext { To ensure accuracy, plug the ordered pair } (-7, 42) ext{ back into both original equations. } ext { ullet } ext { ullet } ext { ullet } ext { For the first equation, } y = -6x ext{ 42 = -6(-7) ext. Since } -6 imes -7 = 42 ext{, this equation holds true. } ext { ullet } ext { ullet } ext { ullet } ext { For the second equation, } y = -8x - 14 ext{ 42 = -8(-7) - 14 ext{. This simplifies to } 42 = 56 - 14 ext{, which further simplifies to } 42 = 42 ext{. Since both equations are satisfied, our solution } (-7, 42) ext{ is correct. } ext {
