Solving Systems With Substitution: A Simple Guide

Hey guys! Today, we're diving into a super cool math topic: solving systems of equations using substitution. If you've ever looked at a system of equations and thought, "Whoa, what am I supposed to do here?!", then this is the video for you. We're going to break down how to use the substitution method to find the solution, and trust me, it's way easier than it looks. We'll be using the example: $\left\{\begin{array}{l} 3 y+x=10 \ y=3 x \end{array}\right.$ This might seem a little intimidating at first, but stick with me, and you'll be a substitution pro in no time. The main idea behind the substitution method is to take one equation and use it to express one variable in terms of another. Then, you substitute that expression into the other equation. This clever little trick reduces a system with two variables into a single equation with just one variable, which is way easier to solve. Think of it like swapping out one piece for another to simplify a puzzle. We'll go step-by-step, making sure you understand each part of the process. We'll even touch on why this method works and when it's most useful. So grab your notebooks, maybe a snack, and let's get started on mastering the substitution method for solving systems of equations! It’s all about making complex problems manageable, and substitution is your best friend for that. We'll make sure that by the end of this, you feel confident tackling any system of equations that comes your way. Remember, math is all about building blocks, and understanding substitution is a huge win in your math journey. So, let's jump right in and conquer this! We've got this, team!

Now, let's get down to business with our specific example: $\left\{\begin{array}{l} 3 y+x=10 \ y=3 x \end{array}\right.$ Our goal here is to find the values of $x$ and $y$ that make both equations true at the same time. The substitution method is perfect for this. You'll notice that the second equation, $y = 3x$, already gives us a nice, clean expression for $y$ in terms of $x$. This is exactly what we want! It's like the problem handed us a golden ticket. Because we know that $y$ is equal to $3x$, we can take this entire expression, $3x$, and substitute it wherever we see $y$ in the first equation. So, in the first equation, $3y + x = 10$, we're going to replace that $y$ with $(3x)$. This gives us: $3(3x) + x = 10$. See what we did there? We swapped $y$ for $3x$. Now, this equation only has one variable, $x$, which we can totally solve. Let's simplify: $3$ times $3x$ is $9x$. So the equation becomes $9x + x = 10$. Combine the $x$ terms: $9x + x$ is $10x$. So, we have $10x = 10$. To find $x$, we just need to divide both sides by $10$. $x = \frac{10}{10}$, which means $x = 1$. Awesome! We've found the value for $x$. But we're not done yet! Remember, we need to find both $x$ and $y$. The next step is to take this value of $x$ that we just found ($x=1$) and plug it back into either of the original equations to find $y$. The second equation, $y = 3x$, is super easy for this. Since $x=1$, we just substitute that in: $y = 3(1)$. That means $y = 3$. So, our solution is $x=1$ and $y=3$. To be super sure, we can always check our answer by plugging these values back into both original equations. For the first equation: $3y + x = 10$. Substitute $x=1$ and $y=3$: $3(3) + 1 = 9 + 1 = 10$. It works! For the second equation: $y = 3x$. Substitute $x=1$ and $y=3$: $3 = 3(1)$. That also works! Since our values satisfy both equations, our solution $(x=1, y=3)$ is correct. High five!

Let's explore a different scenario to really nail down the substitution method for solving systems of equations. Sometimes, the equations aren't as neatly set up as our first example. Imagine you have a system like this: $\left\{\begin{array}{l} 2x + 5y = 1 \ x - 2y = 4 \end{array}\right.$ Here, neither equation directly tells us what $x$ or $y$ equals in terms of the other variable. But don't panic! The substitution method is still our best pal. The first step, as always, is to get one variable isolated in one of the equations. We want to rearrange one of the equations so that it looks like $x = ext{something}$ or $y = ext{something}$. Looking at the second equation, $x - 2y = 4$, it seems pretty easy to get $x$ by itself. All we need to do is add $2y$ to both sides. So, $x - 2y + 2y = 4 + 2y$, which simplifies to $x = 4 + 2y$. Boom! We've now expressed $x$ in terms of $y$. This is our substitution expression. Now, we take this expression for $x$ (which is $4 + 2y$) and substitute it into the other equation, the first one: $2x + 5y = 1$. So, everywhere we see an $x$ in $2x + 5y = 1$, we're going to replace it with $(4 + 2y)$. This gives us: $2(4 + 2y) + 5y = 1$. Now, just like before, we have an equation with only one variable, $y$. Let's solve it! First, distribute the $2$ into the parentheses: $2 imes 4$ is $8$, and $2 imes 2y$ is $4y$. So the equation becomes $8 + 4y + 5y = 1$. Combine the $y$ terms: $4y + 5y$ is $9y$. So we have $8 + 9y = 1$. To isolate the $y$ term, subtract $8$ from both sides: $9y = 1 - 8$, which means $9y = -7$. Finally, divide both sides by $9$ to find $y$: $y = -\frac{7}{9}$. We found our $y$ value! Now, just like in the previous example, we need to find $x$. We take our value for $y$ ($y = -\frac{7}{9}$) and plug it back into the equation where we isolated a variable. Remember we found $x = 4 + 2y$? That's the perfect equation to use. So, $x = 4 + 2(-\frac{7}{9})$. Multiply $2$ by $-\frac{7}{9}$: $2 imes -\frac{7}{9} = -\frac{14}{9}$. Now we have $x = 4 - \frac{14}{9}$. To subtract these, we need a common denominator. $4$ can be written as $\frac{36}{9}$ (since $4 imes 9 = 36$). So, $x = \frac{36}{9} - \frac{14}{9}$. Subtracting the numerators gives us $36 - 14 = 22$. So, $x = \frac{22}{9}$. Our solution is $x = \frac{22}{9}$ and $y = -\frac{7}{9}$. Remember to always check your answer by plugging these values back into both original equations. It’s a bit more work with fractions, but it’s the best way to ensure you got it right! This process, guys, is the heart of the substitution method – isolate, substitute, solve, and then find the other variable. Keep practicing, and it'll become second nature!

So, why bother with the substitution method? When is it your best bet for solving systems of equations? Well, as we've seen, the substitution method shines brightest when one of your equations already has a variable isolated, like $y = 3x$ or $x = 4 + 2y$. If you see that, jump on it! It makes the first step of isolating a variable a total breeze. It's also super useful when the coefficients (the numbers in front of the variables) are simple, especially if one of them is $1$ or $-1$. For example, in the system $\left\{\begin{array}{l} 3 y+x=10 \ y=3 x \end{array}\right.$, the $x$ in the first equation has a coefficient of $1$. We could have isolated $x$ as $x = 10 - 3y$ and substituted that into the second equation. Let's see how that would work, just for kicks. Substituting $x = 10 - 3y$ into $y = 3x$ gives us $y = 3(10 - 3y)$. Distribute the $3$: $y = 30 - 9y$. Add $9y$ to both sides: $y + 9y = 30$, so $10y = 30$. Divide by $10$: $y = 3$. Now, substitute $y=3$ back into $x = 10 - 3y$: $x = 10 - 3(3) = 10 - 9 = 1$. And voilà, we get $x=1, y=3$ again! See? It works no matter which variable you isolate first, as long as you're consistent. This flexibility is a big plus. On the flip side, if none of the variables have a coefficient of $1$ or $-1$, and no variable is already isolated, substitution can still work, but it might involve more fractions early on, like in our second example ($x - 2y = 4$). In cases like that, the elimination method (which we might cover another time!) can sometimes be quicker. But really, guys, the substitution method is a fundamental skill. It’s about strategic replacement to simplify complexity. It reinforces the idea that different algebraic expressions can represent the same value, and we can leverage that equality to solve problems. The more you practice, the better you'll get at spotting the easiest path forward. It’s not just about getting the answer; it’s about understanding the underlying logic and building your problem-solving toolkit. So, embrace the substitution method, practice it, and watch your confidence in tackling systems of equations soar!