Hey guys! Let's dive into the fascinating world of systems of equations and figure out how to determine whether they have one solution, no solution, or infinitely many solutions. We'll tackle a few examples together, making sure to break down each step so it's super clear. Think of it like this: we're detectives, and the equations are clues. Our mission? To uncover the secrets of x and y! So, buckle up, and let's get started!
Understanding Systems of Equations and Their Solutions
First things first, what exactly is a system of equations? Simply put, it's a set of two or more equations containing the same variables. We're usually trying to find values for those variables (like x and y) that make all the equations in the system true at the same time. This is where the magic happens – the solutions! Now, a system of linear equations can behave in one of three ways:
- One Unique Solution: This is the classic scenario. Imagine two lines intersecting at a single point on a graph. That point represents the one and only solution that satisfies both equations.
- No Solution: Picture two parallel lines. They never intersect, which means there's no point (no solution) that works for both equations. These systems are inconsistent.
- Infinitely Many Solutions: This occurs when the two equations essentially represent the same line. They overlap completely, so every point on the line is a solution to both equations. These are dependent systems.
To effectively determine the type of solution a system possesses, we utilize a few key methods. These include the substitution method, the elimination method, and analyzing the slopes and y-intercepts of the equations when they are in slope-intercept form (y = mx + b). For instance, if we have two equations in slope-intercept form, we can quickly assess the situation: if the slopes are different, there is one solution; if the slopes are the same but the y-intercepts are different, there are no solutions; and if both the slopes and y-intercepts are the same, there are infinitely many solutions. The substitution method involves solving one equation for one variable and substituting that expression into the other equation, which can simplify the system and reveal the nature of the solution. The elimination method, also known as the addition method, involves manipulating the equations so that the coefficients of one variable are opposites, allowing us to add the equations together and eliminate that variable, again leading to a clearer understanding of the solution type. Understanding these fundamental concepts and methods equips us to tackle a variety of systems of equations and accurately determine the number of solutions.
Example 1: 3x - 2y = 3, 6x - 4y = 1
Okay, let's tackle our first system: 3x - 2y = 3 and 6x - 4y = 1. The goal here is to figure out if these equations represent intersecting lines (one solution), parallel lines (no solution), or the same line (infinitely many solutions). A super helpful method for this is the elimination method. The elimination method will help us clear out variables and find the possible answer.
To use elimination, we want to make the coefficients of either x or y opposites in the two equations. Notice that if we multiply the first equation by -2, the coefficient of x will become -6, which is the opposite of 6 in the second equation. Let's do that:
-2 * (3x - 2y) = -2 * 3 -6x + 4y = -6
Now we have two equations:
-6x + 4y = -6 6x - 4y = 1
Time for the big moment! Let's add these two equations together. What happens? The -6x and 6x cancel out, and so do the 4y and -4y. This leaves us with:
0 = -5
Whoa! This is a false statement. Zero never equals -5. What does this mean? It means our system of equations has no solution. The lines represented by these equations are parallel and never intersect. Guys, imagine trying to find a meeting point for two trains running on parallel tracks – impossible, right? Same idea here!
Why did this happen? Multiplying the first equation by -2 essentially tried to make it match the second equation, but it couldn't. The left sides looked like they could match (by cancelling terms), but the right sides were fundamentally different, leading to the contradiction. This is a classic sign of a system with no solution.
Example 2: 3x - 5y = 8, 5x - 3y = 2
Alright, next up: 3x - 5y = 8 and 5x - 3y = 2. We're back on the case to determine if this system has one solution, no solutions, or infinitely many. Again, let's bring in the elimination method – it's a powerful tool! With the elimination method, we can find a variable to cancel out and simplify the equations.
This time, it's not as obvious how to make the coefficients opposites with a single multiplication. So, let's get a little creative. Let's aim to eliminate x. We can multiply the first equation by 5 (the coefficient of x in the second equation) and the second equation by -3 (the negative of the coefficient of x in the first equation). This will give us 15x and -15x, which will cancel out when we add the equations.
Here we go:
5 * (3x - 5y) = 5 * 8 => 15x - 25y = 40 -3 * (5x - 3y) = -3 * 2 => -15x + 9y = -6
Now, let's add 'em up:
(15x - 25y) + (-15x + 9y) = 40 + (-6) -16y = 34
Aha! We've isolated y. Let's solve for it:
y = 34 / -16 = -17/8
Fantastic! We've found a value for y. Now, to find x, we can substitute this value back into either of our original equations. Let's use the first one:
3x - 5*(-17/8) = 8 3x + 85/8 = 8 3x = 8 - 85/8 3x = (64 - 85) / 8 3x = -21/8 x = (-21/8) / 3 x = -7/8
Boom! We've found x = -7/8 and y = -17/8. This system has one unique solution. These two lines intersect at the point (-7/8, -17/8). This illustrates that with a bit of algebraic manipulation, systems of equations can be solved to reveal a single, precise solution.
Example 3: 3x + 2y = 8, 4x + 3y = 1
Let's move on to our third system: 3x + 2y = 8 and 4x + 3y = 1. We're still on the hunt to determine the solution type – one, none, or infinite. The elimination method has been our friend, so let's stick with it. We will use the elimination method again to solve this new system of equations.
Looking at the equations, it seems like eliminating either x or y will require a bit of work. Let's aim for eliminating y this time, just to mix things up. We can multiply the first equation by 3 (the coefficient of y in the second equation) and the second equation by -2 (the negative of the coefficient of y in the first equation). This will give us 6y and -6y, which will cancel each other out.
Let's do the multiplications:
3 * (3x + 2y) = 3 * 8 => 9x + 6y = 24 -2 * (4x + 3y) = -2 * 1 => -8x - 6y = -2
Time to add these equations together:
(9x + 6y) + (-8x - 6y) = 24 + (-2) x = 22
Excellent! We've found that x = 22. Now, to find y, we substitute this value back into either of the original equations. Let's use the first one again:
3 * 22 + 2y = 8 66 + 2y = 8 2y = 8 - 66 2y = -58 y = -29
There we have it! We've found x = 22 and y = -29. This system has one unique solution. The lines intersect at the point (22, -29). This reinforces the idea that sometimes, finding the solution requires a bit more algebraic maneuvering, but the elimination method can guide us to the answer.
Example 4: 3x - 5y = 3, 2x - 4y = 2
Now, let's tackle our fourth system of equations: 3x - 5y = 3 and 2x - 4y = 2. Our mission, should we choose to accept it (and we do!), is to determine the number of solutions this system possesses. You guessed it – we're dusting off the elimination method once again. This is a method that lets us eliminate one variable to find the other.
Looking at the equations, we need to decide which variable to eliminate. Neither x nor y has coefficients that are easily made opposites with a single multiplication. So, let's aim to eliminate x. To do this, we can multiply the first equation by 2 (the coefficient of x in the second equation) and the second equation by -3 (the negative of the coefficient of x in the first equation). This will give us 6x and -6x, setting us up for elimination.
Let's multiply:
2 * (3x - 5y) = 2 * 3 => 6x - 10y = 6 -3 * (2x - 4y) = -3 * 2 => -6x + 12y = -6
Time to add the equations together:
(6x - 10y) + (-6x + 12y) = 6 + (-6) 2y = 0
Intriguing! We've found that 2y = 0, which means y = 0. Now, let's substitute this value back into one of our original equations to find x. We will substitute the value we found to solve for the other variable.
Let's use the first equation:
3x - 5 * 0 = 3 3x = 3 x = 1
Hooray! We've found x = 1 and y = 0. This system has one unique solution. The two lines intersect at the point (1, 0). This example highlights how the elimination method can sometimes lead to a straightforward solution, even when the initial equations look a bit complex.
Example 5: 3x - 4y = 2, 6x - 8y = 1
Alright, let's tackle our final system of equations: 3x - 4y = 2 and 6x - 8y = 1. We're in the home stretch, guys! Our goal remains the same: determine whether this system has one solution, no solutions, or infinitely many solutions. And, as our trusty companion, we'll be using the elimination method.
Looking at the equations, we can see a relationship between the coefficients. The coefficients in the second equation (6 and -8) are multiples of the coefficients in the first equation (3 and -4). This is a clue that something interesting might be going on. Let's see if we can make the coefficients of x opposites by multiplying the first equation by -2:
-2 * (3x - 4y) = -2 * 2 -6x + 8y = -4
Now we have two equations:
-6x + 8y = -4 6x - 8y = 1
Let's add them together:
(-6x + 8y) + (6x - 8y) = -4 + 1 0 = -3
Wait a minute! We've landed on another false statement: 0 = -3. This is a clear signal that our system has no solution. The lines represented by these equations are parallel and will never intersect. It is good to remember that parallel lines do not intersect, meaning there is no solution to this system.
What happened here? Multiplying the first equation by -2 made the left side a multiple of the second equation, but the right sides didn't match up. This discrepancy resulted in the contradiction, confirming that the lines are parallel. Remember, if you encounter a false statement like this during the elimination process, you've discovered a system with no solution.
Wrapping Up: Mastering Systems of Equations
So, guys, we've journeyed through several examples of solving systems of equations. We've seen how to use the elimination method to determine whether a system has one unique solution, no solution, or infinitely many solutions. Remember, the key is to manipulate the equations to either eliminate a variable (leading to a solution or a contradiction) or recognize when the equations represent the same line. With practice, you'll become a pro at spotting these different scenarios!
Key takeaways:
- One Solution: Lines intersect at a single point.
- No Solution: Lines are parallel and never intersect (results in a false statement like 0 = -3).
- Infinitely Many Solutions: Equations represent the same line (one equation is a multiple of the other).
Keep practicing, and you'll be solving systems of equations like a boss! You've got this! And remember, math isn't about memorizing steps; it's about understanding the concepts. Once you get the 'why' behind the 'how,' you can tackle any problem. Now go out there and conquer those equations! You're awesome, guys!
