Solve Systems Of Equations: Elimination Method

Hey guys, let's dive into the awesome world of solving systems of equations! Today, we've got a cool problem where we need to figure out how Harvey tackled a system using a clever trick: multiplying each equation by a constant to make those pesky $x$ terms disappear. It's all about making things simpler so we can find that sweet spot where both equations are true. When we talk about solving systems of equations, we're essentially looking for the values of the variables (in this case, $x$ and $y$) that satisfy all the equations in the system simultaneously. This is super important in tons of real-world scenarios, from figuring out the best way to allocate resources to predicting how two objects will interact. The elimination method, which Harvey is using here, is one of the most powerful ways to crack these problems. It relies on a fundamental property of equations: you can multiply an entire equation by any non-zero constant, and it remains equivalent. Think of it like scaling up a recipe; you're not changing the fundamental proportions, just the total amount. Harvey's strategy is to pick multipliers for each equation so that when we add the modified equations together, the $x$ terms cancel each other out, leaving us with a simple equation in just $y$. This is the magic of elimination – it reduces a problem with two unknowns to a problem with just one, which is way easier to solve! We'll walk through his exact steps, showing you how to pick those magic numbers and what the resulting equations look like. Get ready to become a master of equation manipulation!

So, let's look at the system Harvey's working with:

$$7x + 3y = 5$$

$$2x + 5y = -11$$

Harvey's goal is to make the coefficients of the $x$ terms opposites. Right now, we have a $7x$ in the first equation and a $2x$ in the second. To make them cancel out when we add the equations, we need one to be, say, $+14x$ and the other to be $-14x$. How do we achieve this? By multiplying each equation by a carefully chosen constant. For the first equation, $7x + 3y = 5$, we can multiply the entire equation by 2. This gives us:

$$2 * (7x + 3y) = 2 * 5$$

$$14x + 6y = 10$$

See? We haven't changed the truth of the equation, just scaled it up. Now, for the second equation, $2x + 5y = -11$, we need to multiply it by a constant that will make its $x$ term the opposite of $14x$. Since we have $+14x$ in our modified first equation, we need $-14x$ in our modified second equation. To get $-14x$ from $2x$, we need to multiply by $-7$. So, we multiply the entire second equation by $-7$:

$$(-7) * (2x + 5y) = (-7) * (-11)$$

$$-14x - 35y = 77$$

And boom! We have our two new, modified equations:

$$14x + 6y = 10$$

$$-14x - 35y = 77$$

These are the resulting equations after Harvey performs his multiplication. Notice how the coefficients of the $x$ terms are now $14$ and $-14$? They are opposites! This sets us up perfectly for the next step in the elimination method, where we'll add these two equations together. This process of solving systems of equations is all about strategic manipulation, and Harvey's clearly got the hang of it. It's like setting a trap for the $x$ variable, making it disappear so we can focus on $y$. This technique is super useful when you're dealing with equations where the coefficients don't easily divide into each other, making direct substitution a bit more of a headache. By using these multipliers, we create a situation where one variable is guaranteed to be eliminated upon addition or subtraction, streamlining the path to the solution. Keep this in mind as you tackle your own equation puzzles!

The Power of Multiplication in Equation Solving

Alright, let's really unpack why this multiplication step is so crucial when we're solving systems of equations. Think about it, guys: the core principle behind the elimination method is to add or subtract the equations in such a way that one of the variables cancels out completely. If we just tried to add the original equations ($7x + 3y = 5$ and $2x + 5y = -11$), we'd get $(7x+2x) + (3y+5y) = 5 + (-11)$, which simplifies to $9x + 8y = -6$. This doesn't help us eliminate anything; we still have both $x$ and $y$. The magic happens when the coefficients of one variable become additive inverses (like $14$ and $-14$, or $5$ and $-5$). This is precisely what Harvey achieved by multiplying. He didn't just pick numbers randomly; he looked at the coefficients of $x$ (which were $7$ and $2$) and found a common multiple. The least common multiple (LCM) of $7$ and $2$ is $14$. His goal was to transform both $7x$ and $2x$ into terms that would result in $14x$ and $-14x$ (or $-14x$ and $14x$) when the equations were scaled. For the first equation ($7x + 3y = 5$), multiplying by $2$ turned the $7x$ into $14x$. For the second equation ($2x + 5y = -11$), multiplying by $-7$ turned the $2x$ into $-14x$. The choice of $-7$ was strategic: not only did it make the $x$ coefficient the opposite of $14$, but it also ensured that we'd be adding the equations later, which is often simpler and less error-prone than subtraction.

It's also important to remember that every term in the equation must be multiplied by the constant. If Harvey had only multiplied the $x$ term in the first equation by $2$, he would have fundamentally changed the equation and its relationship to the second equation, leading to an incorrect solution. By multiplying $7x + 3y = 5$ by $2$, he got $14x + 6y = 10$. Both the $3y$ and the $5$ had to be multiplied by $2$ as well. Similarly, multiplying $2x + 5y = -11$ by $-7$ resulted in $-14x - 35y = 77$, where the $5y$ became $-35y$ and the $-11$ became $77$. This meticulous application of the multiplication property maintains the equality and balance of each original equation while setting them up for successful elimination. The resulting equations, $14x + 6y = 10$ and $-14x - 35y = 77$, are equivalent to the original equations, meaning they represent the same lines in a graph, but they are now in a form that is much more conducive to finding the solution point where these lines intersect. This is the beauty and power of algebraic manipulation in action, making complex problems solvable with a systematic approach.

The Next Steps: Eliminating $x$ and Solving for $y$

Now that Harvey has masterfully transformed the original system into:

$$14x + 6y = 10$$

$$-14x - 35y = 77$$

He's ready for the most satisfying part of the elimination method: making the $x$ variable vanish! Because the coefficients of $x$ are now perfect opposites ($14$ and $-14$), all we need to do is add the two equations together, term by term. Let's see what happens:

$$(14x + (-14x)) + (6y + (-35y)) = 10 + 77$$

$$0x + (6y - 35y) = 87$$

$$0 - 29y = 87$$

$$-29y = 87$$

And just like that, the $x$ terms are gone! We're left with a super simple equation involving only $y$. To solve for $y$, we just need to divide both sides by $-29$:

$$y = \frac{87}{-29}$$

$$y = -3$$

There we have it! We've found the $y$-coordinate of our solution. This is a huge step, and it was all made possible by carefully choosing those multipliers in the previous step. The process of solving systems of equations often feels like peeling back layers of complexity, and elimination is brilliant at stripping away one variable at a time. The fact that the $x$ terms canceled out so cleanly means our setup was correct. If we had ended up with, say, $5y = 87$, we would know something went wrong in the multiplication stage. But here, $-29y = 87$ is straightforward. This highlights how critical it is to ensure every part of the equation is multiplied by the constant. A slip-up there would propagate through the entire process. This stage is all about reaping the rewards of that initial strategic multiplication. We deliberately engineered the system so that adding them would obliterate the $x$, leaving us with a direct path to $y$. It’s a testament to the elegance of algebra that such a transformation is possible. The beauty of this method is its generality; it works for any system of linear equations where you can find suitable multipliers to create opposite coefficients. This ability to simplify the problem dramatically is why the elimination method is a cornerstone of algebra, particularly when dealing with larger or more complex systems where substitution might become unwieldy. The key takeaway here is that the multiplication step isn't just busywork; it's the engine that drives the elimination process forward, enabling us to isolate and solve for one variable at a time.

Finding $x$ and Completing the Solution

We've successfully found that $y = -3$. But remember, a solution to a system of equations is a pair of values $(x, y)$ that satisfies both original equations. So, we still need to find the value of $x$. Luckily, now that we know $y = -3$, we can substitute this value back into either of the original equations (or even the modified ones, though using the originals is usually safest to avoid introducing errors from the multiplication step). Let's use the first original equation: $7x + 3y = 5$.

Substitute $y = -3$ into the equation:

$$7x + 3(-3) = 5$$

$$7x - 9 = 5$$

Now, this is just a simple one-variable equation for $x$. To solve it, we first add $9$ to both sides:

$$7x = 5 + 9$$

$$7x = 14$$

Finally, divide both sides by $7$:

$$x = \frac{14}{7}$$

$$x = 2$$

So, we've found our $x$ value! The solution to the system of equations is $x = 2$ and $y = -3$, or as an ordered pair, $(2, -3)$.

To be absolutely sure, we can check this solution in the second original equation: $2x + 5y = -11$.

Substitute $x = 2$ and $y = -3$:

$$2(2) + 5(-3) = -11$$

$$4 - 15 = -11$$

$$-11 = -11$$

It checks out! This confirms that our solution $(2, -3)$ is correct. The whole process, starting from Harvey's clever multiplication to create opposite $x$ coefficients, led us directly to this unique solution. Solving systems of equations is all about following these logical steps. The elimination method, particularly when requiring multiplication, demonstrates the power of algebraic transformations. Harvey's approach of multiplying each equation by a constant was the key to unlocking the solution. Without that step, solving might have been much more cumbersome. This problem beautifully illustrates how understanding the properties of equations allows us to manipulate them strategically. Whether you prefer substitution or elimination, the goal is always to simplify the problem and isolate the variables. In this case, Harvey's chosen method proved highly effective, transforming a system that might have seemed initially tricky into one that yielded its solution quite readily. Remember, the ability to perform these operations – multiplication, addition, subtraction, and substitution – is fundamental to mastering algebra and tackling more complex mathematical challenges that you'll encounter down the road. Keep practicing, and you'll be solving systems like a pro in no time!