Solve Equations With Back Substitution: A Step-by-Step Guide

When you're faced with a system of equations, especially one that's already in a nice, organized form, back substitution is your best friend. It's a straightforward method that allows you to solve for variables one by one, starting from the simplest equation and working your way up. This technique is particularly useful when your system is already in what's called row echelon form or upper triangular form, meaning that each equation has one more variable than the equation above it. This structure is exactly what we have in our example:

$\left\{\begin{aligned} a-b+c & =2 \\ 2 b-5 c & =-12 \\ 3 c & =6 \end{aligned}\right.$

See how the first equation has $a$, $b$, and $c$? The second equation only has $b$ and $c$, and the third equation only has $c$? This is the perfect setup for back substitution! It means we can immediately find the value of one variable, and then use that value to find the next, and so on. It's like a mathematical domino effect where each solved variable knocks down the next. We'll go through this specific system step-by-step to show you just how effective and easy back substitution can be. So, grab a pen and paper, and let's dive into solving this system together!

The Power of Back Substitution in Solving Linear Systems

Let's dive deep into the elegance of back substitution, a cornerstone method for solving systems of linear equations, particularly when they're presented in a structured format like upper triangular form. This method is not just about finding a solution; it's about understanding the sequential nature of dependencies within a system. In our given system:

$\left\{\begin{aligned} a-b+c & =2 \\ 2 b-5 c & =-12 \\ 3 c & =6 \end{aligned}\right.$

The beauty of this arrangement is immediately apparent. The third equation, $3c = 6$, is the simplest because it isolates a single variable, $c$. This is where the back substitution process begins. By solving this equation, we get a concrete value for $c$, which then becomes a known quantity. This known value is crucial because it can be substituted into other equations to help solve for other variables. The second equation, $2b - 5c = -12$, contains both $b$ and $c$. Since we will have already determined the value of $c$ from the third equation, we can substitute this value into the second equation. This transforms the second equation into a simpler equation with only one unknown variable, $b$, which we can then solve. Once $b$ is known, we move to the first equation, $a - b + c = 2$. This equation contains $a$, $b$, and $c$. With the values of both $b$ and $c$ now known, substituting them into the first equation will isolate the variable $a$, allowing us to solve for it. This systematic process, moving from the simplest equation to the most complex, is what defines back substitution and makes it an exceptionally efficient method for solving systems of equations that are already organized in a way that facilitates this step-by-step approach. The principle is simple: solve for the last variable, then substitute that value backward into the preceding equation to solve for the second-to-last variable, and continue this process until all variables are determined. It’s a logical progression that demystifies complex systems by breaking them down into manageable steps.

Step 1: Solving for 'c' from the Simplest Equation

Our journey begins with the third equation in the system, which is $3c = 6$. This equation is beautifully simple because it contains only one variable, $c$. The objective here is to isolate $c$ to find its exact numerical value. To do this, we need to get rid of the coefficient '3' that is multiplying $c$. The standard algebraic operation for this is division. We will divide both sides of the equation by 3. Remember, whatever operation you perform on one side of an equation, you must perform the same operation on the other side to maintain the equality.

So, we have:

$\frac{3c}{3} = \frac{6}{3}$

This simplifies to:

$c = 2$

And there you have it! We've successfully found the value of $c$. This is the crucial first step in our back substitution process. Now that we know $c$ is equal to 2, we can use this information to solve for the other variables in the system. Think of $c=2$ as our first solved piece of the puzzle. It's the foundation upon which we'll build the rest of our solution. This single step is often the most straightforward, as it requires no prior knowledge of other variables, just basic arithmetic. The efficiency of back substitution truly shines when you have systems structured like this, where the last equation neatly provides the value of the last variable. It sets a clear path forward for solving the remaining unknowns.

Step 2: Substituting 'c' to Solve for 'b'

Now that we've cracked the code for $c$ and found that $c = 2$, we can move on to the second equation in our system: $2b - 5c = -12$. The goal here is to find the value of $b$. Notice that this equation contains both $b$ and $c$. Since we already know that $c = 2$, we can substitute this value into the equation. This substitution is the heart of the back substitution method – using a known value to simplify another equation.

Let's substitute $c=2$ into the equation:

$2b - 5(2) = -12$

Now, we perform the multiplication:

$2b - 10 = -12$

Our next step is to isolate the term with $b$. To do this, we need to move the '-10' to the other side of the equation. We accomplish this by adding 10 to both sides:

$2b - 10 + 10 = -12 + 10$

This simplifies to:

$2b = -2$

Finally, to solve for $b$, we divide both sides of the equation by 2:

$\frac{2b}{2} = \frac{-2}{2}$

Which gives us:

$b = -1$

Fantastic! We've now found the value of $b$. With $c=2$ and $b=-1$, we're one step away from solving the entire system. This step demonstrates how each solved variable directly contributes to solving the next. It's a chain reaction of calculation, making complex problems manageable.

Step 3: Substituting 'b' and 'c' to Solve for 'a'

We're in the home stretch! We've successfully determined that $c = 2$ and $b = -1$. Now, we turn our attention to the first equation in the system: $a - b + c = 2$. Our final task is to solve for $a$. This equation involves all three variables, $a$, $b$, and $c$. Since we already know the values of $b$ and $c$, we can substitute them into this equation to isolate and solve for $a$.

Let's substitute $b = -1$ and $c = 2$ into the equation:

$a - (-1) + (2) = 2$

First, simplifying the double negative for $b$:

$a + 1 + 2 = 2$

Combine the constant terms on the left side:

$a + 3 = 2$

To isolate $a$, we need to move the '+3' to the other side of the equation. We do this by subtracting 3 from both sides:

$a + 3 - 3 = 2 - 3$

This leaves us with:

$a = -1$

And there we have it! We have found the value of $a$. The complete solution to our system of equations is $a = -1$, $b = -1$, and $c = 2$. The back substitution method has guided us smoothly through each step, revealing the values of all variables sequentially. This method is powerful because it leverages the structure of the system, making the process intuitive and efficient. The satisfaction of solving the entire system by building upon each previous result is a key reward of mastering this technique. It shows the interconnectedness of algebraic expressions and the power of systematic problem-solving.

Verification: Checking Our Solution

It's always a good practice, especially in mathematics, to verify your solution. This means plugging the values we found back into the original equations to ensure they hold true. If our solution is correct, each original equation should result in a true statement.

Our proposed solution is: $a = -1$, $b = -1$, $c = 2$.

Let's check the original equations:

1. Equation 1: $a - b + c = 2$ Substitute the values: $(-1) - (-1) + (2) = 2$ Simplify: $-1 + 1 + 2 = 2$ Result: $2 = 2$. This equation holds true.

2. Equation 2: $2b - 5c = -12$ Substitute the values: $2(-1) - 5(2) = -12$ Simplify: $-2 - 10 = -12$ Result: $-12 = -12$. This equation also holds true.

3. Equation 3: $3c = 6$ Substitute the values: $3(2) = 6$ Simplify: $6 = 6$. This equation is true as well.

Since all three original equations are satisfied with our values of $a$, $b$, and $c$, we can be confident that our solution is correct. This verification step is a crucial part of the problem-solving process, reinforcing the accuracy of our calculations and the effectiveness of the back substitution method. It provides peace of mind and solidifies understanding.

Conclusion: Mastering Back Substitution

We've successfully navigated the system of equations using the back substitution method, starting from the simplest equation and working our way backward to find the values of $a$, $b$, and $c$. We found that $c=2$, then used that to find $b=-1$, and finally used both to find $a=-1$. The verification step confirmed that our solution $(a, b, c) = (-1, -1, 2)$ is indeed correct.

Back substitution is a powerful technique, especially when dealing with systems of linear equations that are already in or can be easily transformed into upper triangular form. It breaks down complex problems into a series of simpler, manageable steps, making it efficient and less prone to errors. Understanding this method not only helps you solve specific problems but also builds a strong foundation for more advanced algebraic concepts.

Remember, the key is to identify the equation with a single variable, solve for it, and then substitute that value into the next equation containing that variable. Repeat this process until all variables are determined. It’s a systematic approach that demystifies algebraic challenges.

For further exploration into solving systems of equations and related mathematical concepts, you might find the resources at Khan Academy and Wolfram MathWorld to be incredibly helpful. They offer detailed explanations, examples, and practice problems that can deepen your understanding.