Solving Systems Of Equations: A Step-by-Step Guide

Hey guys! Ever get stuck with those tricky systems of equations in math? Don't worry, you're not alone! Many students find them a bit challenging, but once you understand the methods, they become much easier to handle. In this guide, we'll break down how to solve systems of equations step by step. We'll tackle three examples with clear explanations, so you can confidently solve similar problems. So, let's dive in and conquer those equations together!

A) Solving -x + 3y = -5 and 2x - 5y = 8

Okay, so let's kick things off with our first system of equations. We have two equations here: -x + 3y = -5 and 2x - 5y = 8. To solve this, we can use a method called elimination. The main goal here is to get rid of one variable so we can solve for the other. Let’s focus on eliminating 'x' first. To do this, we need to make the coefficients of 'x' in both equations the same but with opposite signs. Looking at the equations, we see that the coefficient of 'x' in the first equation is -1, and in the second equation, it’s 2. The least common multiple of 1 and 2 is 2, so we’ll aim to make the coefficient of 'x' in the first equation -2 and keep the coefficient in the second equation as 2.

To achieve this, we'll multiply the entire first equation by 2. Remember, whatever we do to one side of the equation, we must do to the other to keep the equation balanced. So, let's multiply each term in the first equation by 2: 2*(-x) + 2*(3y) = 2*(-5). This simplifies to -2x + 6y = -10. Now, we have a new version of our first equation, which we can use along with the original second equation. Our updated system of equations looks like this: -2x + 6y = -10 and 2x - 5y = 8. Notice that the coefficients of 'x' are now -2 and 2, which are exactly what we wanted!

Now comes the fun part – adding the two equations together. When we add them, the 'x' terms will cancel each other out because -2x + 2x = 0. This leaves us with only the 'y' terms and the constants. So, let’s add the equations: (-2x + 6y) + (2x - 5y) = -10 + 8. Combining like terms, we get (-2x + 2x) + (6y - 5y) = -2. Simplifying further, we have 0x + y = -2, which is just y = -2. Awesome! We’ve found the value of 'y'. Now that we know 'y', we can plug this value back into either of the original equations to solve for 'x'. Let’s use the first original equation, -x + 3y = -5, because it looks a bit simpler. Substitute y = -2 into this equation: -x + 3*(-2) = -5. This simplifies to -x - 6 = -5.

To isolate 'x', we first add 6 to both sides of the equation: -x - 6 + 6 = -5 + 6. This gives us -x = 1. But we want to find 'x', not '-x', so we multiply both sides by -1: (-1)(-x) = (-1)1. This gives us x = -1. So, we’ve found that x = -1. Now we have both 'x' and 'y'. Our solution is x = -1 and y = -2. To make sure we’ve got the correct solution, it’s always a good idea to check our answers by plugging them back into both original equations. Let’s start with the first equation, -x + 3y = -5. Substitute x = -1 and y = -2: -(-1) + 3(-2) = 1 - 6 = -5. This checks out! Now let’s check the second equation, 2x - 5y = 8. Substitute x = -1 and y = -2: 2(-1) - 5*(-2) = -2 + 10 = 8. This also checks out! So, we can confidently say that our solution, x = -1 and y = -2, is correct. Great job!

B) Solving x - 2y = -9 and 1/4x - 1/2y = 10

Alright, let's move on to our second system of equations. This time, we have x - 2y = -9 and 1/4x - 1/2y = 10. This system looks a little trickier because of those fractions in the second equation, but don’t worry, we can handle it! The first thing we’ll want to do is get rid of those fractions. Fractions can make things look more complicated than they are, so clearing them out will simplify our work.

To eliminate the fractions, we need to find the least common denominator (LCD) of the fractions in the second equation. We have 1/4 and 1/2, so the denominators are 4 and 2. The least common multiple of 4 and 2 is 4. So, we'll multiply the entire second equation by 4 to clear the fractions. Remember, we need to multiply every term in the equation by 4 to keep it balanced. So, let's do it: 4*(1/4x) - 4*(1/2y) = 4*10. This simplifies to x - 2y = 40. Now, our second equation looks much cleaner! Our updated system of equations is: x - 2y = -9 and x - 2y = 40. Now that we've cleared the fractions, we can see something interesting. Both equations have the same left-hand side, x - 2y, but different right-hand sides, -9 and 40. This is a clue that something special is going on with this system.

If we try to solve this system using elimination, let's say by subtracting the first equation from the second, we would get: (x - 2y) - (x - 2y) = 40 - (-9). Simplifying this, we get 0 = 49. Wait a minute! That's not true. Zero does not equal 49. This result tells us that the system of equations is inconsistent. In simpler terms, there is no solution that satisfies both equations simultaneously. Think about it graphically: these two equations represent lines, and since they have the same slope (same coefficients for x and y) but different y-intercepts (different constants on the right side), they are parallel lines. Parallel lines never intersect, which means there is no point (x, y) that lies on both lines. So, when you encounter a situation like this, where the variables cancel out and you’re left with a false statement, you can confidently say that the system has no solution. It's a bit like trying to find the intersection of two train tracks that run perfectly parallel – it’s just not going to happen!

C) Solving 5x + 10y = 25 and 15x + 30y = 75

Okay, let's tackle our third and final system of equations: 5x + 10y = 25 and 15x + 30y = 75. At first glance, these equations might seem like a typical system, but let's take a closer look before we jump into solving them. Sometimes, spotting a pattern or a relationship between the equations can save us a lot of time and effort. Notice anything interesting? If we look closely, we can see that the second equation is actually a multiple of the first equation. If we multiply the first equation by 3, we get the second equation: 3 * (5x + 10y) = 3 * 25, which simplifies to 15x + 30y = 75. This means that both equations represent the same line!

When two equations represent the same line, they are called dependent equations. This has a very specific implication for the solution of the system. Instead of having a single, unique solution (like in our first example) or no solution (like in our second example), this system has infinitely many solutions. Think about it: if both equations describe the same line, then every single point on that line is a solution to both equations. There are countless points on a line, so there are countless solutions to this system. So, how do we express these infinitely many solutions? We can't list them all out, obviously! Instead, we express the solutions in terms of one of the variables. This is where things get a little abstract, but hang in there, it's not too complicated. Let's solve one of the equations for one of the variables. We can choose either equation since they are essentially the same. Let’s use the first equation, 5x + 10y = 25, because the numbers are a bit smaller, which might make things easier.

First, let's simplify the equation by dividing every term by 5: (5x)/5 + (10y)/5 = 25/5. This simplifies to x + 2y = 5. Now, let's solve for x. We subtract 2y from both sides: x + 2y - 2y = 5 - 2y. This gives us x = 5 - 2y. This equation tells us how x is related to y. For any value we choose for y, we can plug it into this equation and find the corresponding value for x. Since y can be any real number, there are infinitely many pairs of (x, y) that satisfy this equation. To express the solution set, we write it in terms of y. We can say that the solutions are all the points (x, y) such that x = 5 - 2y. In set notation, this looks like: {(x, y) | x = 5 - 2y}. This is a concise way of saying that the solutions are all pairs of numbers (x, y) where x is equal to 5 minus 2 times y. We could also have solved for y in terms of x, and the solution set would look a little different, but it would still represent the same set of points. The key takeaway here is that when you have dependent equations, you have infinitely many solutions, and you express them by giving a relationship between x and y.

Conclusion

So there you have it! We've tackled three different systems of equations, each with its own unique twist. We’ve seen how to use elimination to find a single solution, how to recognize an inconsistent system with no solutions, and how to handle dependent equations with infinitely many solutions. Solving systems of equations is a fundamental skill in algebra, and it pops up in many different areas of math and science. By understanding these methods, you'll be well-equipped to handle these kinds of problems. Keep practicing, and you'll become a system-solving pro in no time! You've got this!