Solve Linear Equations: X-3y=-2 & X+3y=16

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a classic problem: solving systems of linear equations. You know, those situations where you've got two or more equations, each with a couple of variables, and you need to find the one magical set of values that makes all of them true at the same time. It's like finding the perfect key that unlocks multiple doors simultaneously! Today, we're going to unravel a specific system that might look a little daunting at first glance:

$\begin{array}{l} x-3 y=-2 \\x+3 y=16 \end{array}$

Don't sweat it if you're not a math whiz; we're going to break this down step-by-step, making it super easy to follow. We'll explore different methods, discuss why they work, and help you build confidence in your algebra skills. So, grab your notebooks, maybe a coffee, and let's get ready to become linear equation ninjas!

Understanding Linear Equations and Systems

Alright, let's get our heads around what we're actually dealing with here. A linear equation is basically an equation where the variables (like our 'x' and 'y') are only raised to the power of one. No crazy exponents, no square roots of variables, just simple, straight-line relationships. When we talk about a system of linear equations, we're essentially saying we have two or more of these simple equations, and we're looking for a point (or points) where they all intersect. Think of it like plotting lines on a graph; the solution to the system is the coordinate where all those lines cross paths. For our specific problem, we have:

• Equation 1: $x - 3y = -2$
• Equation 2: $x + 3y = 16$

See how both 'x' and 'y' are just to the power of one? That's what makes them linear. Now, our goal is to find a pair of values for 'x' and 'y' that satisfy both of these equations simultaneously. If we plug these values into Equation 1, the left side should equal the right side (-2). And if we plug the exact same values into Equation 2, the left side should equal the right side (16). It's a bit like a mathematical detective mission, searching for that one crucial clue!

There are a few common strategies we can use to solve these systems. The most popular ones are substitution and elimination (sometimes called the addition method). We might also use graphing, but that can be less precise unless you have super-accurate graphing tools or are dealing with simple integer solutions. We'll focus on substitution and elimination because they're generally more robust and easier to nail down, especially when the solutions aren't nice, round numbers. The cool thing about these methods is that they often lead you to the same answer, reinforcing the idea that there's a unique solution (or sometimes no solution, or infinite solutions – but we'll get to that!). For this particular system, you'll notice something special about the 'y' terms, which might make one method a bit more straightforward. Let's explore them!

Method 1: Elimination - Making Variables Disappear!

The elimination method is a super handy technique when you have linear equations where the coefficients (the numbers in front of the variables) are the same or opposites. In our system:

$\begin{array}{l} x-3 y=-2 \\x+3 y=16 \end{array}$

Check out the 'y' terms: we have '-3y' in the first equation and '+3y' in the second. Bingo! These are opposites. This is exactly what we want for elimination. The idea is to add (or subtract) the two equations together in a way that one of the variables cancels out completely, leaving us with a single equation with only one variable. This is why it's called elimination – we're eliminating a variable!

So, let's add Equation 1 and Equation 2 directly. We'll line them up vertically:

  (x - 3y) = -2
+ (x + 3y) = 16
----------------

Now, we add column by column:

• x terms: $x + x = 2x$
• y terms: $-3y + 3y = 0y = 0$. See? The 'y' variable is eliminated!
• Constant terms: $-2 + 16 = 14$

Putting it all together, we get:

$2x + 0 = 14$

Which simplifies to:

$2x = 14$

This is now a super simple linear equation with just one variable, 'x'. To solve for 'x', we just need to divide both sides by 2:

$x = 14 / 2$

$x = 7$

Awesome! We've found the value of 'x'. But remember, a solution to a system of equations is a pair of values (x, y). We're only halfway there. Now that we know $x = 7$, we can use this value and substitute it back into either of the original equations to find the value of 'y'. Let's pick Equation 1 (but Equation 2 would work just as well):

$x - 3y = -2$

Substitute $x = 7$:

$7 - 3y = -2$

Now, we need to isolate 'y'. First, subtract 7 from both sides:

$-3y = -2 - 7$

$-3y = -9$

Finally, divide both sides by -3 to get 'y' by itself:

$y = -9 / -3$

$y = 3$

So, using the elimination method, we found that $x = 7$ and $y = 3$. The solution to our system of linear equations is the point (7, 3).

But hold up, let's just double-check our work, because that's what mathematicians do! We need to make sure that this pair (7, 3) satisfies both original equations.

• Check with Equation 1: $x - 3y = -2$ Substitute $x=7, y=3$: $7 - 3(3) = 7 - 9 = -2$. Yep, it works!

• Check with Equation 2: $x + 3y = 16$ Substitute $x=7, y=3$: $7 + 3(3) = 7 + 9 = 16$. Yep, it works too!

Since our solution (7, 3) works for both equations, we can be super confident that this is the correct answer. The elimination method was pretty slick here because those 'y' coefficients were already opposites, making it a direct path to solving for 'x'.

Method 2: Substitution - Replacing Variables!

Now, let's explore another powerful technique: the substitution method. This method is particularly useful when one of the variables in one of the equations has a coefficient of 1 or -1, or when it's already isolated. For our system:

$\begin{array}{l} x-3 y=-2 \\x+3 y=16 \end{array}$

Notice that in both equations, the 'x' term has a coefficient of 1. This makes it super easy to isolate 'x' in either equation. Let's choose Equation 1 and solve for 'x':

$x - 3y = -2$

To get 'x' by itself, we just need to add $3y$ to both sides:

$x = -2 + 3y$

Now, here's the core idea of substitution: we take this expression for 'x' (which is $-2 + 3y$) and substitute it into the other equation (Equation 2). We're essentially replacing the 'x' in Equation 2 with this equivalent expression. This will give us an equation with only one variable, 'y', which we can then solve.

Equation 2 is: $x + 3y = 16$

Substitute $x = -2 + 3y$ into Equation 2:

$(-2 + 3y) + 3y = 16$

Now, let's simplify and solve for 'y'. First, combine the 'y' terms on the left side:

$-2 + (3y + 3y) = 16$

$-2 + 6y = 16$

Next, we want to get the term with 'y' by itself. Add 2 to both sides of the equation:

$6y = 16 + 2$

$6y = 18$

Finally, divide both sides by 6 to find the value of 'y':

$y = 18 / 6$

$y = 3$

Fantastic! We've found $y = 3$. Just like with the elimination method, this is only half the solution. Now we need to find the value of 'x'. We can do this by taking our value of $y = 3$ and substituting it back into either of the original equations, OR, even easier, into the expression we found for 'x' earlier:

$x = -2 + 3y$

Substitute $y = 3$ into this expression:

$x = -2 + 3(3)$

$x = -2 + 9$

$x = 7$

And there you have it! Using the substitution method, we again arrive at the solution $x = 7$ and $y = 3$. The solution to the system is the point (7, 3).

It's great that both methods yielded the same result. This is a key principle in algebra: different valid methods should lead you to the same correct answer. If you get different answers using elimination and substitution, it's a strong sign that you might have made a calculation error somewhere along the line. Always double-check your steps!

Let's quickly recap the substitution process:

1. Isolate a variable: Choose one equation and solve it for one variable (e.g., solve for 'x' or 'y').
2. Substitute: Take the expression you found for that variable and substitute it into the other equation.
3. Solve for the remaining variable: You'll now have an equation with only one variable. Solve it.
4. Back-substitute: Take the value you just found and plug it back into either of the original equations (or the isolated expression) to find the value of the other variable.

The substitution method is especially powerful when dealing with systems where one variable is already isolated or has a coefficient of 1, as it minimizes the initial steps. It's a core technique for understanding more complex algebraic manipulations.

Graphical Interpretation: Where Lines Meet

So, we've solved our system of linear equations using both elimination and substitution, and we consistently found the solution to be $(7, 3)$. But what does this actually mean in a visual sense? This is where the graphical interpretation comes into play. Remember how we said linear equations represent straight lines on a graph? Well, a system of two linear equations represents two lines. The solution to the system is the point where these two lines intersect.

Let's take our original equations and think about them as lines:

• Line 1: $x - 3y = -2$
• Line 2: $x + 3y = 16$

If we were to graph these two lines on the same coordinate plane, they would cross each other at exactly one point. That single point of intersection would have the coordinates $(7, 3)$. This visual confirmation is a fantastic way to understand what a solution represents. It's the unique common point that belongs to both lines.

To visualize this, we could rewrite each equation in slope-intercept form ($y = mx + b$), where 'm' is the slope and 'b' is the y-intercept. Let's do that:

For Line 1 ($x - 3y = -2$):

1. Subtract 'x' from both sides: $-3y = -x - 2$
2. Divide by -3: $y = (-1/-3)x + (-2/-3)$
3. Simplify: $y = (1/3)x + (2/3)$ So, Line 1 has a slope of $1/3$ and a y-intercept of $2/3$.

For Line 2 ($x + 3y = 16$):

1. Subtract 'x' from both sides: $3y = -x + 16$
2. Divide by 3: $y = (-1/3)x + (16/3)$ So, Line 2 has a slope of $-1/3$ and a y-intercept of $16/3$.

Notice that the slopes are different ($1/3$ and $-1/3$). This is crucial! When two lines have different slopes, they are guaranteed to intersect at exactly one point. If the slopes were the same, the lines would either be parallel (no intersection, no solution) or be the exact same line (infinite intersections, infinite solutions). Since our slopes are different, we know there's a unique solution, which we found to be $(7, 3)$.

If you were to plot these lines accurately, you would see them crossing precisely at the point where $x$ is 7 and $y$ is 3. This graphical interpretation helps solidify the abstract algebraic solutions into a concrete geometric understanding. It's a beautiful connection between algebra and geometry!

When Solutions Don't Exist (or Exist Infinitely)

While our system happily provided a single, clean solution $(7, 3)$, it's important for guys to know that not all systems of linear equations behave this way. There are three possibilities for the solutions of a system of two linear equations in two variables:

1. One Unique Solution: This is what we just saw. The lines intersect at a single point. This happens when the slopes of the two lines are different.
2. No Solution: The lines are parallel and never intersect. This happens when the lines have the same slope but different y-intercepts. Algebraically, when you try to solve the system using elimination or substitution, you'll end up with a false statement, like $0 = 5$. This contradiction indicates that there's no pair of (x, y) values that can satisfy both equations.
3. Infinitely Many Solutions: The two equations represent the exact same line. This happens when the lines have the same slope and the same y-intercept. Algebraically, when you try to solve the system, you'll end up with a true statement, like $0 = 0$ or $5 = 5$. This means any point on the line is a solution, and since there are infinite points on a line, there are infinitely many solutions.

For our specific system:

$\begin{array}{l} x-3 y=-2 \\x+3 y=16 \end{array}$

We found unique values for x and y ($x=7, y=3$). This confirms that our system falls into the first category: it has one unique solution. The graphical representation of these two lines clearly shows them intersecting at a single point. Understanding these three possibilities gives you a complete picture when analyzing systems of linear equations.

Conclusion: Mastering Linear Equations

So there you have it, folks! We've successfully navigated the system of linear equations:

$\begin{array}{l} x-3 y=-2 \\x+3 y=16 \end{array}$

Using two powerful methods, elimination and substitution, we discovered the unique solution to be $x=7$ and $y=3$, represented as the coordinate point $(7, 3)$. We also touched upon the graphical interpretation, where this solution is the point of intersection for the two lines represented by the equations.

Solving systems of linear equations is a fundamental skill in mathematics, with applications ranging from economics and engineering to computer graphics and beyond. Whether you're facing homework problems, standardized tests, or real-world challenges, mastering these techniques will give you a significant edge.

Remember the key takeaways:

• Elimination works wonders when coefficients are opposites or can easily be made opposites.
• Substitution is efficient when a variable is already isolated or has a simple coefficient.
• Always check your solution by plugging the values back into the original equations.
• Be aware that systems can have one solution, no solution, or infinite solutions.

Keep practicing these methods, and don't be afraid to experiment with different approaches. The more you solve, the more intuitive these processes will become. Keep up the great work, and we'll catch you in the next article on Plastik Magazine!