Elimination Method For Solving Systems Of Equations

When you're faced with a system of linear equations, like the one presented:

$\begin{array}{l} 4 x-15 y=-2 \\ x-3 y=1 \end{array}$

there are several methods you can use to find the solution, which is the point (or points) where the lines represented by these equations intersect. Today, we're going to dive deep into the elimination method. This technique is particularly powerful when the coefficients of the variables in the equations are easily manipulated to cancel each other out. It's like a strategic game of removing unwanted pieces to reveal the hidden solution. We'll explore how it works, why it's effective, and walk through solving our example problem step-by-step. Get ready to become a master of elimination!

Understanding the Elimination Method

The core idea behind the elimination method is to manipulate one or both equations in a system so that when you add or subtract the equations, one of the variables is eliminated, or 'canceled out.' This leaves you with a single equation containing only one variable, which is much simpler to solve. Once you've found the value of that variable, you can substitute it back into one of the original equations to find the value of the other variable. It's a systematic approach that relies on the properties of equality: whatever you do to one side of an equation, you must do to the other to maintain balance. This ensures that the solution you find is valid for both original equations.

To effectively use elimination, you want the coefficients of either the $x$ or $y$ terms to be opposites (e.g., $5y$ and $-5y$) or the same (e.g., $2x$ and $2x$). If they are opposites, you add the equations together. If they are the same, you subtract one equation from the other. Often, you'll need to multiply one or both equations by a constant to achieve this. For instance, if you have $2x + 3y = 7$ and $x + y = 3$, you might multiply the second equation by $-2$ to get $-2x - 2y = -6$. Then, adding this modified second equation to the first equation would eliminate $x$. This strategic multiplication is a key part of mastering the elimination technique, allowing you to tackle a wider range of systems.

Step-by-Step Elimination

Let's apply the elimination method to our system:

$\begin{array}{l} 4 x-15 y=-2 \\ x-3 y=1 \end{array}$

Our goal is to make the coefficients of either $x$ or $y$ opposites or identical. Looking at the equations, the coefficients of $y$ are $-15$ and $-3$. The coefficients of $x$ are $4$ and $1$. It seems easiest to manipulate the second equation to match the $y$ coefficients. If we multiply the entire second equation by $5$, the $y$ coefficient will become $-15$, which is the same as in the first equation.

So, let's multiply the second equation ($x - 3y = 1$) by $5$:

$5 * (x - 3y) = 5 * 1$

This gives us:

$5x - 15y = 5$

Now, we have a new system of equations:

$\begin{array}{l} 4 x-15 y=-2 \\ 5 x-15 y=5 \end{array}$

Notice that both equations now have a $-15y$ term. To eliminate $y$, we need to subtract one equation from the other. Let's subtract the first equation from the second equation to keep things positive:

$(5x - 15y) - (4x - 15y) = 5 - (-2)$

Distribute the negative sign:

$5x - 15y - 4x + 15y = 5 + 2$

Combine like terms:

$(5x - 4x) + (-15y + 15y) = 7$

$x + 0y = 7$

$x = 7$

We've successfully eliminated $y$ and found that $x = 7$. Now, we need to find the value of $y$. We can substitute $x=7$ into either of the original equations. Let's use the second original equation, $x - 3y = 1$, because it looks simpler:

$7 - 3y = 1$

Now, we solve for $y$. Subtract $7$ from both sides:

$-3y = 1 - 7$

$-3y = -6$

Divide both sides by $-3$:

$y = -6 / -3$

$y = 2$

So, the solution to the system of equations is $x=7$ and $y=2$, which can be written as the ordered pair $(7, 2)$.

Verification of the Solution

It's always a good practice to check your solution by substituting the values of $x$ and $y$ back into both of the original equations. This helps ensure you haven't made any errors during the elimination or substitution steps.

Let's check with the first original equation: $4x - 15y = -2$

Substitute $x=7$ and $y=2$:

$4(7) - 15(2) = -2$

$28 - 30 = -2$

$-2 = -2$

This equation holds true.

Now, let's check with the second original equation: $x - 3y = 1$

Substitute $x=7$ and $y=2$:

$7 - 3(2) = 1$

$7 - 6 = 1$

$1 = 1$

This equation also holds true. Since our solution $(7, 2)$ satisfies both original equations, we can be confident that it is the correct solution to the system.

When Elimination Shines

The elimination method truly shines when the coefficients of the variables are already opposites or can be easily made into opposites or identical values. Consider a system like:

$\begin{array}{l} 2x + 3y = 7 \\ 4x - 3y = 5 \end{array}$

Here, the $y$ coefficients are already opposites ($+3y$ and $-3y$). Simply adding these two equations together will immediately eliminate $y$:

$(2x + 3y) + (4x - 3y) = 7 + 5$

$6x = 12$

$x = 2$

Then, substituting $x=2$ back into either original equation will give you $y$. This is much quicker than, say, using substitution, where you might have to deal with fractions if you try to isolate a variable.

Another scenario where elimination is advantageous is when dealing with larger numbers or when substitution would involve complicated fractions. For example:

$\begin{array}{l} 5x + 7y = 29 \\ 3x - 2y = 7 \end{array}$

To eliminate $x$, you could multiply the first equation by $3$ and the second by $-5$ (or vice versa) to get coefficients of $15x$ and $-15x$. To eliminate $y$, you could multiply the first equation by $2$ and the second by $7$ to get $14y$ and $-14y$. While it requires more multiplication steps, the resulting numbers are often manageable, and you avoid the potential for calculation errors that can arise from complex fractional substitutions. The key is to choose the variable to eliminate and the multipliers that will simplify the process the most.

Comparison with Other Methods

While elimination is a powerful tool, it's helpful to understand how it compares to other methods for solving systems of equations, such as substitution and graphing. Each method has its own strengths and ideal use cases.

Graphing: This method involves plotting both equations on a coordinate plane. The point where the lines intersect is the solution. Graphing is excellent for visualizing the solution and understanding the geometric interpretation of a system of equations. However, it's often not practical for finding exact solutions, especially if the intersection point has non-integer coordinates or if the lines are very close together. Precision can be a significant issue with the graphing method.

Substitution: In the substitution method, you solve one equation for one variable and then substitute that expression into the other equation. This is particularly useful when one of the equations is already solved for a variable (e.g., $y = 2x + 1$) or when it's easy to isolate a variable without creating complex fractions. It's a direct way to reduce the system to a single equation. However, if isolating a variable results in fractions, the subsequent calculations can become cumbersome and prone to error.

Elimination: As we've seen, the elimination method excels when coefficients can be easily matched or made opposites. It often leads to cleaner calculations when fractions are involved in the original equations or when substitution would require significant algebraic manipulation. It's a systematic approach that can be very efficient for many types of systems. The choice of method often comes down to the specific form of the equations presented. Sometimes, one method will be significantly more straightforward than others.

Conclusion

Mastering the elimination method provides you with a robust and efficient way to solve systems of linear equations. By strategically manipulating equations to eliminate one variable, you can simplify the problem into a single-variable equation, making it much easier to find the values of $x$ and $y$. Remember to always verify your solution by plugging the values back into the original equations. This ensures accuracy and builds confidence in your results. Whether you're tackling homework problems or real-world applications where systems of equations are used, the elimination method is an indispensable skill in your mathematical toolkit.

For further exploration into algebraic techniques and solving equations, you can find valuable resources at Khan Academy. They offer comprehensive lessons and practice exercises on a wide range of mathematical topics.