Solving Systems Of Equations With Elimination Method

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of mathematics, specifically tackling a common challenge: solving systems of equations using the elimination method. Our focus will be on a problem presented by André, who's trying to find the solution to the following system:

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

André's goal is to eliminate one of the variables to simplify the system and find the values of 'a' and 'b'. Let's break down how the elimination method works and see which resulting equations André could achieve.

Understanding the Elimination Method

The elimination method, sometimes called the addition method, is a super useful technique for solving systems of linear equations. The main idea is to add or subtract one equation from the other in a way that eliminates one of the variables. This leaves you with a simpler equation containing only one variable, which you can then solve. Once you have the value of one variable, you can substitute it back into one of the original equations to find the value of the other variable.

In André's system, we have two equations:

Equation 1: $3a + 6b = 12$ Equation 2: $-3a + 6b = -12$

When looking at these equations, notice the coefficients of the 'a' terms. We have a $+3a$ in Equation 1 and a $-3a$ in Equation 2. These are perfect for elimination! If we add the two equations together, the 'a' terms will cancel each other out, leaving us with an equation solely in terms of 'b'.

Let's perform the addition:

$(3a + 6b) + (-3a + 6b) = 12 + (-12)$

Combining like terms:

$(3a - 3a) + (6b + 6b) = 12 - 12$

$0a + 12b = 0$

$12b = 0$

So, one possible resulting equation when André eliminates 'a' is $12b = 0$. This is a key step in solving the system!

Exploring Other Elimination Possibilities

Now, what if André wanted to eliminate the 'b' variable instead? Looking at the original equations:

Equation 1: $3a + 6b = 12$ Equation 2: $-3a + 6b = -12$

Both equations have a $+6b$ term. To eliminate 'b', we would need to subtract one equation from the other. Let's try subtracting Equation 2 from Equation 1:

$(3a + 6b) - (-3a + 6b) = 12 - (-12)$

Distribute the negative sign:

$3a + 6b + 3a - 6b = 12 + 12$

Combine like terms:

$(3a + 3a) + (6b - 6b) = 24$

$6a + 0b = 24$

$6a = 24$

In this case, if André eliminated 'b' by subtracting Equation 2 from Equation 1, the resulting equation would be $6a = 24$. This is another valid outcome of the elimination process.

What if André subtracted Equation 1 from Equation 2?

$(-3a + 6b) - (3a + 6b) = -12 - 12$

$-3a + 6b - 3a - 6b = -24$

$(-3a - 3a) + (6b - 6b) = -24$

$-6a + 0b = -24$

$-6a = -24$

This is also a possible resulting equation if the subtraction order was reversed.

Analyzing the Given Options

Now, let's look at the options provided in the question:

• $12b = 0$: As we saw, this is a direct result of adding the two original equations to eliminate 'a'. So, this is a possible resulting equation.
• $6a = 0$: This equation is not directly obtainable by simply adding or subtracting the original equations as they are. If we were to manipulate the equations further after eliminating 'b' (e.g., if $6a = 24$, then $a=4$), this doesn't directly lead to $6a=0$ as an intermediate step in elimination.
• $-6a = 0$: Similar to the previous option, this is not a direct outcome of the initial elimination steps on the given system.

The Solution Process

Let's continue solving the system using the result $12b = 0$ that we found by eliminating 'a'.

From $12b = 0$, we can easily solve for 'b': $b = \frac{0}{12}$ $b = 0$

Now that we have the value of 'b', we can substitute it back into either of the original equations to find 'a'. Let's use Equation 1:

$3a + 6b = 12$

Substitute $b=0$:

$3a + 6(0) = 12$

$3a + 0 = 12$

$3a = 12$

Now, solve for 'a':

$a = \frac{12}{3}$

$a = 4$

So, the solution to the system of equations is $a=4$ and $b=0$. This means the point $(4, 0)$ is where the lines represented by these equations intersect.

Confirming the Elimination Result

The question asks which could be the resulting equations when André eliminates one of the variables. Based on our work, when André eliminates 'a' by adding the two equations, he gets $12b = 0$. This is one of the options provided.

If he were to eliminate 'b' by subtracting the equations, he would get $6a = 24$ or $-6a = -24$. While $6a=24$ isn't listed as an option in that exact form, $12b=0$ is listed and is a direct result of the elimination process.

Why the Other Options Are Less Likely (in this direct elimination step)

• $6a=0$ and $-6a=0$: These equations would imply that $a=0$. If $a=0$, substituting into the original equations gives:
  • Equation 1: $3(0) + 6b = 12 ightarrow 6b = 12 ightarrow b=2$
  • Equation 2: $-3(0) + 6b = -12 ightarrow 6b = -12 ightarrow b=-2$ Since we get different values for 'b', $a=0$ is not the correct value for 'a' in this system. Therefore, equations leading directly to $a=0$ through simple elimination are not correct outcomes for this specific problem.

Conclusion

When André applies the elimination method to the given system of equations, the most straightforward and correct resulting equation among the choices is $12b=0$. This is achieved by adding the two original equations together, which effectively eliminates the variable 'a'.

Keep practicing, guys! The elimination method is a fundamental skill in algebra, and with a little practice, you'll be solving systems of equations like a pro. Let us know in the comments if you have any questions or other math problems you'd like us to break down!

Stay tuned for more math adventures here at Plastik Magazine!