Solving Systems Of Equations By Elimination

Solving Systems of Equations by Elimination: A Step-by-Step Guide

When you're faced with a system of equations, like the one presented with $3x - y = 28$ and $3x + y = 14$, one of the most effective methods to find the solution (the point where both equations intersect) is the elimination method. This technique is particularly handy when the coefficients of one of the variables are opposites or the same. Our goal in elimination is to strategically add or subtract the equations to eliminate one of the variables, leaving you with a simpler equation that can be solved for the remaining variable. Let's dive into how we can conquer this system using elimination, making complex math feel a whole lot simpler. We'll explore the nuances of this method, offering tips and tricks to make sure you're not just solving problems, but truly understanding the underlying mathematical principles at play.

Understanding the Elimination Method

The core idea behind the elimination method is to manipulate a system of linear equations so that when you combine them (usually by adding or subtracting), one of the variables cancels out. Imagine you have two lines on a graph, and you're looking for the exact spot where they cross. The elimination method is a way to find those coordinates without necessarily graphing them. It's all about making one of the variables disappear, at least temporarily, so you can isolate the other. This is achieved by ensuring that the coefficients of either the 'x' terms or the 'y' terms are either identical or additive inverses (opposites). For example, if one equation has a $+5y$ and another has a $-5y$, adding those equations will make the 'y' term vanish. Similarly, if both equations have a $2x$, subtracting one equation from the other will eliminate the 'x' term. The beauty of this method lies in its directness and efficiency, especially when dealing with equations that are already neatly aligned for cancellation. We'll be focusing on how to apply this to our specific problem, $3x - y = 28$ and $3x + y = 14$, breaking down each step to ensure clarity and confidence in your problem-solving abilities. We aim to provide a comprehensive understanding, moving beyond rote memorization to a deeper appreciation of algebraic manipulation.

Applying Elimination to Our System

Now, let's put the elimination method into action with our specific system: $3x - y = 28$ and $3x + y = 14$. Take a close look at the coefficients of the 'y' terms. In the first equation, we have $-y$, and in the second, we have $+y$. Notice how these are perfect opposites! This is exactly what we need for elimination. When we add these two equations together, the $-y$ and $+y$ will cancel each other out, leaving us with an equation solely in terms of 'x'. This is the magic of elimination – it simplifies the problem dramatically. We simply stack the equations vertically and perform addition:

$(3x - y) + (3x + y) = 28 + 14$

Combine the 'x' terms: $3x + 3x = 6x$. Combine the 'y' terms: $-y + y = 0$. Combine the constants: $28 + 14 = 42$.

This gives us our simplified equation: $6x = 42$. Solving this for 'x' is straightforward. Divide both sides by 6: $x = 42 / 6$, which means $x = 7$. We've successfully eliminated 'y' and found the value of 'x'! This initial step is crucial, and recognizing when elimination is possible—by looking for matching or opposite coefficients—is key to mastering this technique. The elegance of this process is that it transforms a two-variable problem into a one-variable problem in a single step, paving the way for a clear path to the solution. We'll continue by using this value of 'x' to find 'y', completing the solution set for our system.

Finding the Value of the Other Variable

We've successfully used the elimination method to find that $x = 7$. Our next crucial step is to find the corresponding value of 'y'. To do this, we simply substitute the value of 'x' we found (which is 7) into either of the original equations. Both equations will yield the same result for 'y' if our value of 'x' is correct. Let's choose the first equation: $3x - y = 28$.

Substitute $x = 7$: $3(7) - y = 28$

Now, simplify: $21 - y = 28$

To solve for 'y', we need to isolate it. Subtract 21 from both sides of the equation:

$-y = 28 - 21$ $-y = 7$

Finally, to get the value of 'y' (not $-y$), multiply both sides by -1:

$y = -7$

So, the solution for 'y' in this equation is -7. To double-check our work, let's substitute $x = 7$ into the second original equation, $3x + y = 14$:

$3(7) + y = 14$ $21 + y = 14$

Subtract 21 from both sides:

$y = 14 - 21$ $y = -7$

As you can see, both equations give us $y = -7$. This confirms that our value for 'x' was correct, and we have found the complete solution to the system of equations. The ordered pair representing the solution is $(7, -7)$. This process of substitution and verification is vital to ensure accuracy. It reinforces the concept that the solution must satisfy all equations within the system simultaneously, showcasing the integrity of algebraic manipulation.

Verifying the Solution

Once you've solved a system of equations, whether through elimination, substitution, or graphing, the final and perhaps most important step is verification. This is your chance to ensure that your answer is absolutely correct and that you haven't made any calculation errors along the way. For our system, we found the solution to be the ordered pair $(7, -7)$. To verify this, we must substitute these values for 'x' and 'y' back into both of the original equations. If both equations hold true with these values, then our solution is confirmed. Let's start with the first equation: $3x - y = 28$.

Substitute $x = 7$ and $y = -7$: $3(7) - (-7) = 28$

Perform the multiplication: $21 - (-7) = 28$

Subtracting a negative is the same as adding a positive: $21 + 7 = 28$

$28 = 28$

This equation is true! Now, let's check the second equation: $3x + y = 14$.

Substitute $x = 7$ and $y = -7$: $3(7) + (-7) = 14$

Perform the multiplication: $21 + (-7) = 14$

Add the terms: $21 - 7 = 14$

$14 = 14$

This equation is also true! Since the ordered pair $(7, -7)$ satisfies both original equations, we can confidently state that this is the correct solution to the system. This verification step is not just a formality; it's a fundamental part of the scientific method applied to mathematics – testing your hypothesis (your solution) against the established facts (the original equations). It builds confidence in your mathematical skills and reduces the likelihood of submitting an incorrect answer. Always take the time to verify; it's a small effort that guarantees accuracy and solidifies your understanding.

When Elimination Might Be Tricky

While the elimination method is powerful, there are instances where it's not immediately obvious how to apply it. This often happens when the coefficients of neither the 'x' nor the 'y' terms are the same or opposites. In such cases, you'll need to perform an extra step before you can eliminate a variable: you must multiply one or both equations by a constant. The goal here is to adjust the coefficients so that they become opposites or the same. For example, if you have the system:

$2x + 3y = 10$ $x + y = 3$

Neither the 'x' nor the 'y' coefficients match or are opposites. However, notice that if we multiply the second equation by -2, the 'x' coefficients will become opposites ($2x$ and $-2x$). Let's see how that works:

Multiply the second equation by -2: $-2(x + y) = -2(3)$ $-2x - 2y = -6$

Now our system looks like this:

$2x + 3y = 10$ $-2x - 2y = -6$

Now, we can add these two equations together. The $2x$ and $-2x$ will cancel out, leaving us to solve for 'y'. After finding 'y', we can substitute it back into one of the original equations to find 'x'. If the coefficients don't easily become opposites, you might need to multiply both equations. For instance, to eliminate 'y' in the system:

$3x + 2y = 8$ $2x + 5y = 12$

You might multiply the first equation by 5 and the second by -2 (or vice versa) to make the 'y' coefficients opposites (10y and -10y). This preparatory step requires careful attention to arithmetic, as multiplying an entire equation by a constant can introduce new numbers that need to be managed accurately. However, once this adjustment is made, the standard elimination process can proceed smoothly. This adaptability is what makes the elimination method a versatile tool in your algebraic arsenal.

Conclusion

Mastering the elimination method is a significant step in becoming proficient with systems of equations. We've seen how, with our example system $3x - y = 28$ and $3x + y = 14$, the elimination process was remarkably straightforward due to the opposite coefficients of 'y'. By adding the equations, we swiftly eliminated 'y', allowing us to solve for $x = 7$. Subsequently, substituting this value back into either original equation yielded $y = -7$. The final verification step confirmed that our solution, $(7, -7)$, correctly satisfies both equations. Remember, the key to successful elimination lies in observing the coefficients and either adding or subtracting equations to cancel out one of the variables. If the coefficients aren't immediately ready for cancellation, don't hesitate to multiply one or both equations by appropriate constants to set them up for elimination. This method, when applied diligently, offers an efficient and elegant pathway to solving systems of linear equations. For further exploration into algebraic techniques and problem-solving strategies, I recommend visiting Khan Academy's comprehensive resources on algebra. Their interactive lessons and practice problems can provide additional insights and reinforce your understanding of these fundamental mathematical concepts.