How Many Solutions Do These Systems Of Equations Have?

Hey guys! Today, we're diving deep into the fascinating world of systems of linear equations. You know, those pairs of equations that look like they're having a little chat? We're going to figure out just how many solutions each of these systems can have. It's like being a detective, but instead of clues, we're looking for points where lines intersect (or don't!). Let's break down three specific systems and see what secrets they hold.

System (1): A Unique Intersection Point

Alright, let's kick things off with our first system of equations:

(1) $: \begin{cases} 4 x-2 y=4 \\ 5 y+6 x=5 \end{cases}$

When we're trying to find the number of solutions for a system of linear equations, we're essentially asking how many points $(x, y)$ satisfy both equations simultaneously. For systems of two linear equations with two variables, there are three possibilities: one unique solution, no solutions, or infinitely many solutions. The first system we're looking at is designed to give us a unique solution. This means the graphs of these two lines will intersect at exactly one point. To determine this without actually solving for $x$ and $y$, we can compare the slopes and y-intercepts of the lines represented by these equations. Let's rewrite both equations in the standard slope-intercept form, $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

For the first equation, $4x - 2y = 4$, we can isolate $y$:

$-2y = -4x + 4$

Divide by -2:

$y = 2x - 2$

So, for the first line, the slope ($m_1$) is 2 and the y-intercept ($b_1$) is -2.

Now, let's rearrange the second equation, $5y + 6x = 5$, to also get it into $y = mx + b$ form:

$5y = -6x + 5$

Divide by 5:

y = -rac{6}{5}x + 1

For the second line, the slope ($m_2$) is -rac{6}{5} and the y-intercept ($b_2$) is 1.

Now, let's compare our findings. We have $m_1 = 2$ and m_2 = -rac{6}{5}. Since the slopes are different (2 \neq -rac{6}{5}), the lines are not parallel and they will definitely intersect at some point. Furthermore, since the y-intercepts are also different ($b_1 = -2$ and $b_2 = 1$), they don't start at the same point, which is expected if they intersect only once. The fact that the slopes are different is the key indicator. When the slopes of two lines in a system of equations are different, those lines must cross at exactly one point. This single point of intersection represents the unique solution to the system. This system is consistent and independent. The lines are not identical (which would lead to infinite solutions), nor are they parallel and distinct (which would lead to no solutions). So, for system (1), we can confidently say there is one unique solution. It's pretty neat how we can figure this out just by looking at the slopes, right? We don't even need to go through the whole substitution or elimination process to know the number of solutions. That's a handy shortcut when you just need the big picture!

System (2): Infinite Solutions or None?

Let's move on to our second system, which presents a different kind of scenario:

(2) $: \begin{cases} 4 x-2 y=4 \\ 8 x-4 y=8 \end{cases}$

This system looks a bit different, and it hints at a special case. When we're analyzing the number of solutions, systems like this often fall into the categories of either no solutions or infinitely many solutions. Remember, no solutions mean the lines are parallel and distinct (they never meet), and infinitely many solutions mean the lines are actually the same line (they overlap everywhere!). To figure out which one it is, we'll again use our trusty method of converting the equations to slope-intercept form, $y = mx + b$.

Let's take the first equation: $4x - 2y = 4$. We'll isolate $y$ just like before:

$-2y = -4x + 4$

Divide by -2:

$y = 2x - 2$

So, the first line has a slope ($m_1$) of 2 and a y-intercept ($b_1$) of -2.

Now, let's work on the second equation: $8x - 4y = 8$. Get $y$ by itself:

$-4y = -8x + 8$

Divide by -4:

$y = 2x - 2$

Wow, check this out! The second line also has a slope ($m_2$) of 2 and a y-intercept ($b_2$) of -2.

What does this mean? It means both equations represent the exact same line! The second equation is just the first equation multiplied by 2. $2 imes (4x - 2y = 4)$ gives us $8x - 4y = 8$. Since both equations describe the identical line, every single point on that line is a solution to both equations.

When this happens, we say the system has infinitely many solutions. Any $(x, y)$ pair that satisfies $y = 2x - 2$ will be a solution. Think about it: if you pick any $x$, calculate the corresponding $y$, that pair will work for both the original equations because they are fundamentally the same equation.

This type of system is called consistent and dependent. Consistent because there are solutions (in fact, tons of them!), and dependent because one equation can be derived from the other. It's a bit like saying the same thing twice in different words – you're still conveying the same meaning. So, for system (2), the answer is infinitely many solutions. It's a cool contrast to the first system, showing how slight changes in coefficients can lead to drastically different outcomes in terms of solutions.

System (3): The Case of Parallel Lines

Finally, let's tackle system (3). This one is designed to show us the other extreme scenario:

(3) $: \begin{cases} 4 x-2 y=4 \\ 8 x-4 y=10 \end{cases}$

We're back to comparing slopes and y-intercepts to determine the number of solutions. As we saw with system (2), if the slopes are the same but the y-intercepts are different, we end up with parallel lines. Parallel lines, by definition, never intersect. If they never intersect, they can't share any common points, which means the system has no solutions. Let's verify this by converting our equations to the $y = mx + b$ format.

Starting with the first equation: $4x - 2y = 4$. We already solved this one in the previous examples, so we know:

$y = 2x - 2$

This line has a slope ($m_1$) of 2 and a y-intercept ($b_1$) of -2.

Now, let's rearrange the second equation: $8x - 4y = 10$.

$-4y = -8x + 10$

Divide by -4:

$y = \frac{-8}{-4}x + \frac{10}{-4}$

$y = 2x - \frac{5}{2}$

For the second line, the slope ($m_2$) is 2 and the y-intercept ($b_2$) is -rac{5}{2} (or -2.5).

Now, let's compare. We have $m_1 = 2$ and $m_2 = 2$. The slopes are identical! This tells us that the lines are parallel. However, their y-intercepts are different: $b_1 = -2$ and b_2 = -rac{5}{2}. Since the lines have the same slope but different y-intercepts, they are parallel and distinct. They run alongside each other but will never cross paths.

Because these lines never intersect, there is no point $(x, y)$ that can satisfy both equations simultaneously. Therefore, this system has no solutions. This situation is described as an inconsistent system. It's inconsistent because there's no value of $x$ and $y$ that can make both statements true at the same time. It's like asking someone to be in two different places at once – it's impossible!

So, to recap for system (3): same slope, different y-intercepts means no solutions. It’s the complete opposite of having infinitely many solutions, where the lines are identical. It’s crucial to be able to distinguish between these cases because it tells us so much about the relationship between the equations we're working with. Pretty cool, right? We've now seen all three possibilities: one solution, infinite solutions, and no solutions, all by comparing the slopes and intercepts of the lines.