Solving Linear Systems Finding The Value Of A

Hey guys! Today, we're diving into the exciting world of linear systems and tackling a problem where we need to find the value of a specific variable. Linear systems are sets of equations where we have multiple variables, and our goal is to find the values of those variables that satisfy all the equations simultaneously. It might sound intimidating, but trust me, it's like piecing together a puzzle! So, let's get started and break down this problem step by step.

Understanding the Problem

First, let's take a good look at the linear system we're dealing with. We have three equations, each involving the variables a, b, and c. Our mission, should we choose to accept it (and we do!), is to find the value of a. Here's the system:

a - b + c = -6
b - c = 5
2a - 2c = 4

Before we jump into solving, let's take a moment to appreciate what we have. We've got three equations and three unknowns – a, b, and c. This is a classic setup for solving a linear system. There are several methods we can use, like substitution, elimination, or even matrices (if you're feeling fancy!). We are going to use the substitution method in order to solve this problem. It involves solving one equation for one variable and substituting that expression into other equations.

Why is Solving Linear Systems Important?

Now, you might be wondering, "Why bother with all this?" Well, linear systems pop up in all sorts of real-world situations! From engineering and physics to economics and computer science, they're used to model relationships between different quantities. For instance, we can use them to analyze electrical circuits, balance chemical equations, or even predict the trajectory of a rocket. So, mastering the art of solving linear systems is a valuable skill to have in your mathematical toolkit. It's like having a superpower for tackling complex problems!

The Strategy: Substitution Method

Okay, let's talk strategy. For this particular problem, the substitution method seems like a good fit. The second equation, b - c = 5, is relatively simple and already has b almost isolated. So, let's start by solving that equation for b. This means we want to get b all by itself on one side of the equation. We can do this by adding c to both sides:

b - c + c = 5 + c
b = 5 + c

Ta-da! We've got b in terms of c. Now, we can substitute this expression for b into the first equation. This will eliminate b from the first equation, leaving us with an equation involving only a and c. Think of it like a mathematical magic trick – we're making a variable disappear!

Diving Deeper into Substitution

The substitution method is a powerful technique, but it's not just about blindly plugging things in. It's about strategically choosing which equation to solve for which variable. We picked the second equation because it was easy to isolate b. If we had tried to solve the first equation for a, for example, we would have ended up with a more complicated expression involving both b and c. So, a little bit of planning can save us a lot of work in the long run. It's like choosing the right tool for the job – a screwdriver for a screw, a wrench for a bolt, and the substitution method for a linear system with a conveniently isolated variable!

Applying the Substitution

Now comes the fun part – actually doing the substitution! We're going to take our expression for b (b = 5 + c) and plug it into the first equation (a - b + c = -6). This means wherever we see a b in the first equation, we're going to replace it with (5 + c). Let's do it:

a - (5 + c) + c = -6

Notice the parentheses around (5 + c). This is super important! We need to make sure we distribute the negative sign correctly. Think of it like this: we're subtracting the entire expression (5 + c), not just the 5. So, let's distribute that negative sign:

a - 5 - c + c = -6

Now, we can simplify the equation by combining the c terms. We have a -c and a +c, which cancel each other out. It's like they're playing tug-of-war and the rope ends up right in the middle:

a - 5 = -6

We're getting closer! We've eliminated b and c from the first equation, leaving us with a simple equation involving only a. This is exactly what we wanted! We're like mathematical detectives, piecing together the clues to solve the mystery.

The Power of Simplification

Simplifying equations is a crucial skill in algebra. It's like cleaning up a messy room – once everything is organized, it's much easier to see what you have and where you need to go. In this case, simplifying our equation allowed us to isolate a and move closer to our goal. Simplification often involves combining like terms, distributing, and using the order of operations (PEMDAS/BODMAS). It's the unsung hero of equation solving!

Solving for a

We're in the home stretch now! We have the equation a - 5 = -6, and our goal is to isolate a. To do this, we need to get rid of the -5 on the left side. The opposite of subtracting 5 is adding 5, so let's add 5 to both sides of the equation:

a - 5 + 5 = -6 + 5

The -5 and +5 on the left side cancel each other out, leaving us with just a:

a = -1

We did it! We found the value of a! It's like reaching the top of a mountain after a long climb – the view is amazing!

Double-Checking Our Work

But hold on, we're not quite done yet. It's always a good idea to double-check our work, especially in math. We can do this by plugging our value for a back into the original equations and seeing if they hold true. This is like verifying our solution with a trusted source – we want to make sure our answer is rock solid.

Using the Third Equation

But wait! We still haven't used the third equation, 2a - 2c = 4. This equation provides us with a direct relationship between a and c, which is perfect for finding the value of c. We already know a = -1, so let's substitute that into the third equation:

2(-1) - 2c = 4
-2 - 2c = 4

Now, we can solve for c. First, let's add 2 to both sides:

-2 - 2c + 2 = 4 + 2
-2c = 6

Then, let's divide both sides by -2:

-2c / -2 = 6 / -2
c = -3

Great! We found that c = -3. Now, we have values for both a and c. We're like expert codebreakers, deciphering the hidden values within the system!

Finding b

We're on a roll! Now that we know c = -3, we can use our earlier expression for b (b = 5 + c) to find the value of b:

b = 5 + (-3)
b = 2

So, we've found that b = 2. We now have values for all three variables: a = -1, b = 2, and c = -3. We're like mathematical superheroes, saving the day by solving the linear system!

Verification

Let's verify the solutions.

Verify Equation 1

Substitute the values of a, b, and c into the first equation:

a - b + c = -6
-1 - 2 + (-3) = -6
-6 = -6

The first equation holds true.

Verify Equation 2

Substitute the values of b and c into the second equation:

b - c = 5
2 - (-3) = 5
5 = 5

The second equation also holds true.

Verify Equation 3

Substitute the values of a and c into the third equation:

2a - 2c = 4
2(-1) - 2(-3) = 4
-2 + 6 = 4
4 = 4

The third equation holds true as well. Our solutions are correct!

The Answer

We've successfully navigated the world of linear systems and found the value of a! Looking back at the multiple-choice options:

• A. -1
• B. 1
• C. 2
• D. -3

We can confidently say that the correct answer is A. -1.

Key Takeaways

Solving linear systems is a fundamental skill in mathematics with wide-ranging applications. The substitution method, which we used in this problem, is a powerful technique for solving systems where one variable can be easily isolated. Remember to always double-check your work to ensure accuracy, and don't be afraid to break down complex problems into smaller, manageable steps. You got this! And that's how we conquer linear systems, guys! Keep practicing, and you'll become a math whiz in no time!