Mastering The Addition Method For Systems Of Equations

Welcome, math enthusiasts! Today, we're diving deep into a powerful technique for solving systems of linear equations: the addition method. This method, also known as the elimination method, is a fantastic tool when you want to find the exact point where two lines intersect. We'll be tackling a specific example to illustrate just how effective it is, transforming a pair of equations into a single, solvable equation. So, grab your notebooks, and let's get started on unraveling the mystery behind these intersecting lines and finding that unique solution.

Understanding the Addition Method

The addition method is an algebraic technique used to solve a system of two linear equations with two variables. The core idea behind this method is to manipulate one or both equations in such a way that when you add them together, one of the variables cancels out (is eliminated). This leaves you with a single equation containing only one variable, which you can then solve. Once you have the value of one variable, you can substitute it back into either of the original equations to find the value of the other variable. It's a systematic approach that guarantees accuracy and efficiency, especially when dealing with equations that aren't immediately amenable to substitution. The beauty of the addition method lies in its simplicity once you grasp the fundamental principle of elimination. It’s like a well-choreographed dance where terms align perfectly to lead you to the solution.

Why Use the Addition Method?

While other methods like substitution exist, the addition method shines when the coefficients of one or both variables in the two equations are opposites (like 3 and -3) or can be easily made into opposites. For instance, if you have a "3y" in one equation and a "-3y" in the other, adding them directly eliminates "y". If the coefficients aren't opposites, you can multiply one or both equations by a specific number to create opposite coefficients. This strategic multiplication ensures that when you add the equations, the desired variable vanishes. This makes it a preferred method when dealing with equations where isolating a variable for substitution would be cumbersome or lead to fractions. It streamlines the process, making complex systems feel more manageable and reducing the potential for calculation errors. Think of it as preparing the battlefield for a decisive elimination, ensuring the variables are set up for a clean sweep.

Step-by-Step Solution

Let's tackle the system of equations you've presented:

$$\left\{\begin{array}{l} 3 x-3 y=7 \ 9 x+6 y=16 \end{array}\right.$$

Our goal is to eliminate either $x$ or $y$. Looking at the coefficients, the $y$ terms (-3y and 6y) seem like a good target. To make them opposites, we can multiply the first equation by 2. This will turn the -3y into -6y, which is the opposite of +6y.

• Step 1: Multiply the first equation by 2.

  $$2(3x - 3y) = 2(7)$$

  $$6x - 6y = 14$$

• Step 2: Rewrite the system with the modified first equation.

  $$\left\{\begin{array}{l} 6x - 6y = 14 \ 9x + 6y = 16 \end{array}\right.$$

• Step 3: Add the two equations together. Notice how the $y$ terms cancel out:

  $$(6x - 6y) + (9x + 6y) = 14 + 16$$

  $$6x + 9x - 6y + 6y = 30$$

  $$15x = 30$$

• Step 4: Solve for x.

  $$x = \frac{30}{15}$$

  $$x = 2$$

• Step 5: Substitute the value of x back into one of the original equations. Let's use the first equation ($3x - 3y = 7$):

  $$3(2) - 3y = 7$$

  $$6 - 3y = 7$$

• Step 6: Solve for y.

  $$-3y = 7 - 6$$

  $$-3y = 1$$

  $$y = -\frac{1}{3}$$

• Step 7: Write the solution as an ordered pair. The solution is $(2, -\frac{1}{3})$.

Verification

To ensure our solution is correct, we should always verify it by substituting the values of $x$ and $y$ back into both original equations. This step is crucial and can save you a lot of headaches by catching potential errors early on.

• Check in the first equation: $3x - 3y = 7$

  $$3(2) - 3(-\frac{1}{3}) = 6 - (-1) = 6 + 1 = 7$$

  This equation holds true!

• Check in the second equation: $9x + 6y = 16$

  $$9(2) + 6(-\frac{1}{3}) = 18 + (-2) = 18 - 2 = 16$$

  This equation also holds true!

Since our values for $x$ and $y$ satisfy both equations, we can be confident that our solution, $(2, -\frac{1}{3})$, is indeed correct. This verification process is a fundamental part of algebraic problem-solving, reinforcing the understanding that a valid solution must satisfy all conditions of the system.

When Coefficients Don't Easily Match

What happens if the coefficients aren't as cooperative as in our example? For instance, consider a system like:

$$\left\{\begin{array}{l} 2x + 3y = 5 \ 4x + 5y = 9 \end{array}\right.$$

Here, neither the $x$ coefficients (2 and 4) nor the $y$ coefficients (3 and 5) are opposites, nor can they be made opposites by multiplying just one equation. In such cases, the addition method still works, but it requires a bit more finesse. You'll need to multiply both equations by carefully chosen numbers so that the coefficients of one variable become opposites. For example, to eliminate $x$ in the system above, you could multiply the first equation by -2 and the second equation by 1. This would give you:

$$\left\{\begin{array}{l} -4x - 6y = -10 \ 4x + 5y = 9 \end{array}\right.$$

Now, when you add these modified equations, the $x$ terms cancel out:

$$(-4x - 6y) + (4x + 5y) = -10 + 9$$

$$-y = -1$$

$$y = 1$$

Once you have $y=1$, you can substitute it back into either of the original equations to find $x$. This principle of strategic multiplication is what makes the addition method a versatile and powerful tool in your algebraic arsenal. It allows you to systematically break down complex systems into simpler, solvable parts, transforming seemingly daunting problems into manageable steps. The key is to identify a common multiple for the coefficients you wish to eliminate and then multiply each equation accordingly to achieve the desired opposites.

Conclusion

The addition method is a fundamental technique for solving systems of linear equations. By strategically manipulating equations to eliminate one variable, you can efficiently find the unique solution where the lines represented by the equations intersect. Remember to always verify your solution by substituting the values back into the original equations to ensure accuracy. With practice, this method becomes an intuitive and powerful tool in your mathematical toolkit.

For further exploration and practice on solving systems of equations, you can visit Khan Academy or Paul's Online Math Notes. These resources offer comprehensive explanations and a wealth of practice problems to solidify your understanding.