Solve Linear Equations: $2a - 3c = -6$ & $a + 2c = 11$

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Hey guys! Ever stared at a system of linear equations and felt like you needed a secret decoder ring? Well, fear not! Today, we're diving deep into a classic problem that'll have you solving for $(a, c)$ like a pro. We've got a couple of equations here: $2a - 3c = -6$ and $a + 2c = 11$. Our mission, should we choose to accept it, is to find the magical pair of values for $a$ and $c$ that make both of these equations true at the same time. It's like trying to find the perfect handshake that satisfies two different sets of rules! We'll explore the most common methods to crack this code, from the trusty substitution method to the slick elimination technique. By the end of this, you'll be able to tackle similar problems with confidence and impress your friends (or at least, your math teacher!). So grab your thinking caps, and let's get these equations sorted!

The Substitution Method: A Step-by-Step Breakdown

Alright, let's talk about the substitution method, one of the most straightforward ways to solve this system of linear equations. The main idea here is to isolate one variable in one of the equations and then substitute that expression into the other equation. It’s like swapping out a piece in a puzzle to make the whole picture fit. Let's start with our equations:

1. $2a - 3c = -6$
2. $a + 2c = 11$

Looking at these, the second equation ($a + 2c = 11$) seems like the easiest one to work with to isolate a variable. Why? Because the coefficient of $a$ is just 1. We can easily get $a$ by itself. So, let's rearrange the second equation to solve for $a$:

$a = 11 - 2c$

Now that we have an expression for $a$, we can take this entire expression ($11 - 2c$) and plug it into the first equation wherever we see '$a$'. This is the substitution step, the heart of this method. So, the first equation ($2a - 3c = -6$) becomes:

$2(11 - 2c) - 3c = -6$

See what we did there? We replaced '$a$' with '$11 - 2c$'. Now, we have an equation with only one variable, $c$, which is awesome! Let's simplify and solve for $c$:

First, distribute the 2:

$22 - 4c - 3c = -6$

Combine the $c$ terms:

$22 - 7c = -6$

Now, we want to get the $c$ term by itself. Subtract 22 from both sides:

$-7c = -6 - 22$

$-7c = -28$

Finally, divide by -7 to find the value of $c$:

$c = \frac{-28}{-7}$

$c = 4$

Boom! We've found our value for $c$. It's 4. But we're not done yet! We still need to find the value of $a$. Remember that expression we got when we isolated $a$ from the second equation? $a = 11 - 2c$. We can use this to find $a$ now that we know $c=4$:

$a = 11 - 2(4)$

$a = 11 - 8$

$a = 3$

So, our solution is $a = 3$ and $c = 4$. This means the solution to the system of equations is the pair $(3, 4)$. Pretty neat, right? The substitution method turned a two-variable problem into a one-variable problem, which is much easier to handle. We just need to be careful with our algebra! Always double-check your work, especially when distributing and combining terms. This method is super reliable and a fundamental tool in your math arsenal.

The Elimination Method: A Sleek Alternative

Next up, let's explore the elimination method, another powerful technique for solving systems of linear equations. The goal here is to manipulate one or both equations so that when you add or subtract them, one of the variables cancels out (is eliminated). It’s like setting up a mathematical duel where one contestant is eliminated, leaving the other to claim victory. This method can be particularly efficient when the coefficients of one of the variables are the same or opposites. Let's look at our system again:

1. $2a - 3c = -6$
2. $a + 2c = 11$

We want to make the coefficients of either $a$ or $c$ match (or be opposites) in both equations. If we want to eliminate $a$, we could multiply the second equation by 2. If we want to eliminate $c$, we could multiply the first equation by 2 and the second equation by 3. Let's try eliminating $a$. To do this, we need the coefficient of $a$ in the second equation to be $-2$ so it cancels out with the $2a$ in the first equation when we add them. We can achieve this by multiplying the entire second equation by $-2$:

$-2 * (a + 2c) = -2 * 11$

This gives us:

$-2a - 4c = -22$

Now we have a new system:

1. $2a - 3c = -6$
2. $-2a - 4c = -22$

Notice how the coefficients of $a$ are now $2$ and $-2$. They are opposites! This means if we add the two equations together, the $a$ terms will disappear:

$(2a - 3c) + (-2a - 4c) = -6 + (-22)$

$2a - 3c - 2a - 4c = -6 - 22$

Combine like terms. The $2a$ and $-2a$ cancel each other out:

$-7c = -28$

This is the exact same equation for $c$ we got using the substitution method! Solving for $c$ by dividing both sides by $-7$:

$c = \frac{-28}{-7}$

$c = 4$

Awesome! We found $c=4$. Now, just like with substitution, we need to find $a$. We can substitute $c=4$ back into either of the original equations. Let's use the second original equation ($a + 2c = 11$) because it looks simpler:

$a + 2(4) = 11$

$a + 8 = 11$

Subtract 8 from both sides:

$a = 11 - 8$

$a = 3$

So, once again, we arrive at the solution $a = 3$ and $c = 4$. The elimination method allowed us to neatly remove one variable, making the problem much more manageable. It's a fantastic technique, especially when you have coefficients that are easily made into opposites or identical numbers.

Verifying Your Solution: Does it Really Work?

So, we've found our potential solution: $(a, c) = (3, 4)$. But how do we know for sure it's correct? The best way is to verify it by plugging these values back into both of the original equations. If the equations hold true, then we've nailed it! This step is super important, guys, as it catches any little slip-ups in our calculations.

Let's check the first equation: $2a - 3c = -6$

Substitute $a=3$ and $c=4$:

$2(3) - 3(4) = -6$

$6 - 12 = -6$

$-6 = -6$

Looks good! The first equation checks out.

Now, let's check the second equation: $a + 2c = 11$

Substitute $a=3$ and $c=4$:

$3 + 2(4) = 11$

$3 + 8 = 11$

$11 = 11$

Perfect! The second equation also holds true. Since $(3, 4)$ satisfies both equations simultaneously, we can be absolutely confident that this is the correct solution to the system of linear equations. This verification step is your best friend in math – always use it!

Which Option is the Winner?

We've worked through the problem using both the substitution and elimination methods, and consistently arrived at the solution $(a, c) = (3, 4)$. Now, let's compare this to the options provided:

A. $\left(-\frac{76}{7}, \frac{17}{7}\right)$ B. $(3,-4)$ C. $(3,4)$ D. $\left(\frac{87}{7},-\frac{5}{7}\right)$

Our calculated solution, $(3, 4)$, directly matches option C. So, that's our winner! It's always satisfying to see your hard-earned answer line up perfectly with one of the choices. This confirms our calculations and our understanding of the methods used. Remember, whether you prefer substitution or elimination, the key is to apply the steps logically and accurately, and always, always verify your answer. Keep practicing, and you'll be solving these systems in your sleep!