No Solution Systems: Linear Combinations Explained

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a common head-scratcher: systems of equations that have no solution. You know, those tricky problems where lines are parallel and never, ever meet. We're going to break down a specific example and explore what it means to find a linear combination of such a system. This stuff might sound intimidating, but trust me, by the end of this article, you'll be feeling like a math whiz!

So, let's get straight to it. We're looking at this system:

$$\left\{\begin{array}{l} \frac{2}{3} x+\frac{5}{2} y=15 \\ 4 x+15 y=12 \end{array}\right.$$

The question asks: Which equation could represent a linear combination of the system? And we're given some options, but the crucial piece of info here is that this system has no solution. What does that even mean in the grand scheme of things? When we graph these two equations, they represent lines. If a system has no solution, it means these lines are parallel. They have the same slope but different y-intercepts, so they'll never intersect. No intersection point means no (x, y) pair satisfies both equations simultaneously. This is super important because understanding why a system has no solution is key to understanding linear combinations of that system. A linear combination of two equations is essentially creating a new equation by multiplying each of the original equations by some constants and then adding them together. Think of it like mixing two ingredients – you're combining them in a specific ratio to get something new. The cool thing is, if the original equations represent parallel lines, any valid linear combination of them will also represent a line parallel to the original two. It won't suddenly magically create a solution where there wasn't one. The new line will just be another parallel line, continuing that theme of never meeting. So, when we look for a linear combination, we're essentially looking for an equation that fits this pattern: it's derived from the original two, and it maintains that characteristic of parallelism. Let's get into how we actually find one of these equations.

Now, how do we actually find a linear combination? The general form of a linear combination of our two equations is c_1(rac{2}{3} x+rac{5}{2} y=15) + c_2(4 x+15 y=12), where $c_1$ and $c_2$ are constants (numbers). Our goal is to pick values for $c_1$ and $c_2$ and then simplify the resulting equation. The beauty of linear combinations is that any solution to the original system would also be a solution to the combined equation. However, since this system has no solution, any linear combination we create will also have no solution. The problem boils down to figuring out which of the given options could be formed by such a combination.

To make this easier, let's first try to simplify our original equations. Multiplying the first equation by 6 (the least common multiple of 3 and 2) gets rid of the fractions:

6 imes (rac{2}{3} x+rac{5}{2} y=15) ightarrow 4x + 15y = 90

Now, look at our system again:

$$\left\{\begin{array}{l} 4x + 15y = 90 \\ 4x + 15y = 12 \end{array}\right.$$

See what happened? Both equations now have the exact same left-hand side ($4x + 15y$), but different right-hand sides (90 and 12). This is the hallmark of a system with no solution – parallel lines with different y-intercepts. If we try to solve this, we could subtract the second equation from the first:

$(4x + 15y) - (4x + 15y) = 90 - 12$ $0 = 78$

This is a contradiction, which confirms our system has no solution.

Now, let's think about the linear combination. Let's say we multiply the first original equation by $c_1$ and the second by $c_2$. We get:

c_1(rac{2}{3} x+rac{5}{2} y) = 15c_1 $c_2(4 x+15 y) = 12c_2$

Adding these together: (c_1 rac{2}{3} x + c_2 4x) + (c_1 rac{5}{2} y + c_2 15y) = 15c_1 + 12c_2

(rac{2}{3} c_1 + 4 c_2) x + (rac{5}{2} c_1 + 15 c_2) y = 15c_1 + 12c_2

This is the general form of a linear combination. The key is that the ratio of the coefficients of x and y on the left side should be the same as in the original equations (which is 4/15 after clearing fractions) for it to represent a line parallel to the originals. Or, more precisely, if we look at the original fractions, the ratio of x coefficients should be proportional to the ratio of y coefficients.

Let's try a simple linear combination. What if we multiply the first original equation by 3 and the second by -1? This is just an example to see what we get:

3 imes (rac{2}{3} x+rac{5}{2} y=15) ightarrow 2x + rac{15}{2} y = 45 $-1 imes (4 x+15 y=12) ightarrow -4x - 15y = -12$

Adding these two new equations: (2x - 4x) + (rac{15}{2} y - 15y) = 45 - 12 -2x - rac{15}{2} y = 33

This equation, -2x - rac{15}{2} y = 33, is a linear combination of the original system. Notice that if you were to clear the fractions here by multiplying by 2, you'd get $-4x - 15y = 66$. This new equation has the same left-hand side as our second original equation ($4x+15y=12$) but a different right-hand side. This is consistent because our original system, after clearing fractions, was $4x+15y=90$ and $4x+15y=12$. Any linear combination will result in an equation of the form $4x+15y = ext{some number}$.

The specific linear combination given as option A is rac{4}{3} x. This isn't even an equation in the standard form $Ax + By = C$. It's just a term. For an equation to represent a linear combination of a system of two linear equations, it must involve both variables, x and y, and be in the form $Ax + By = C$. Options that are just single terms like rac{4}{3} x or equations that don't maintain the parallel nature of the lines (i.e., they would intersect the original lines) are not valid linear combinations that represent a line parallel to the originals.

Let's re-examine the structure of a linear combination. We are looking for an equation of the form $A'x + B'y = C'$ where A' = rac{2}{3} c_1 + 4 c_2 and B' = rac{5}{2} c_1 + 15 c_2 and $C' = 15 c_1 + 12 c_2$. A key property of parallel lines is that the ratio of the coefficients of x and y is the same. In our simplified original system ($4x+15y=90$ and $4x+15y=12$), the ratio of the x-coefficients is $4/4 = 1$, and the ratio of the y-coefficients is $15/15 = 1$. This means the coefficients of x and y are identical. So, any linear combination that results in a line parallel to these will also have identical coefficients for x and y (or coefficients that are proportional in the same way).

Consider the equation rac{4}{3} x. This is not a valid equation representing a line in the standard form $Ax+By=C$ derived from a system of two linear equations. A linear combination of two equations in x and y will always result in another equation in x and y. Therefore, rac{4}{3} x cannot be a linear combination of the given system.

Let's think about what kind of equation could be a linear combination. We saw that multiplying the first equation by 6 gave us $4x + 15y = 90$. The second equation is $4x + 15y = 12$.

Let's try to create a combination that results in something like rac{4}{3} x IF it were a valid equation form. But it isn't. The question implies we're looking for an equation of the form $Ax+By=C$.

Suppose we are given options for a linear combination. For example, let's say one of the options was $8x + 30y = 24$. How would we check if this is a linear combination? We'd see if we can find $c_1, c_2$ such that:

rac{2}{3} c_1 + 4 c_2 = 8 rac{5}{2} c_1 + 15 c_2 = 30 $15 c_1 + 12 c_2 = 24$

Let's simplify the second equation. Divide by 15: rac{1}{6} c_1 + c_2 = 2, so c_2 = 2 - rac{1}{6} c_1. Now substitute this into the first equation: rac{2}{3} c_1 + 4 (2 - rac{1}{6} c_1) = 8 rac{2}{3} c_1 + 8 - rac{4}{6} c_1 = 8 rac{2}{3} c_1 - rac{2}{3} c_1 = 0 $0 = 0$

This means that the first two equations are dependent. They don't give us unique values for $c_1$ and $c_2$. This makes sense because the coefficients of x and y in $8x+30y$ are double the coefficients of $4x+15y$. So, if we chose $c_1=0$, we would need $4c_2 = 8$ (so $c_2=2$) and $15c_2 = 30$ (so $c_2=2$). This fits! Now let's check the third condition with $c_1=0$ and $c_2=2$: $15(0) + 12(2) = 24$. This is true! So, $8x + 30y = 24$ is a linear combination of the original system. Notice that $8x+30y=24$ is just $2 imes (4x+15y=12)$. This is a linear combination where $c_1=0$ and $c_2=2$.

What if the option was rac{4}{3} x? This equation only has an x term. A linear combination of two equations in x and y will always result in an equation that, in general, contains both x and y terms, unless the coefficients of one of the variables happen to cancel out perfectly for all possible combinations, which is not the case here.

The core takeaway: A linear combination of a system of equations creates a new equation that is a valid consequence of the original equations. For systems with no solution (parallel lines), any linear combination will result in another equation representing a parallel line. The equation rac{4}{3} x is not in the form $Ax+By=C$ and cannot be derived as a meaningful linear combination of the given system. It's crucial to look for an equation that has both x and y terms and whose coefficients maintain the parallel relationship, or simply check if the equation can be formed by summing multiples of the original equations. Since the provided option is rac{4}{3} x, and it doesn't fit the structure of a linear combination derived from a system of two linear equations in x and y, it's the one that could not represent a linear combination. The question asks which could represent one, implying we need to find a valid derived equation. If the options were like $8x+30y=24$, that would be a valid linear combination. Without seeing the other options, we can definitively say that rac{4}{3} x alone cannot be a linear combination of the given system. It's possible the question or options are presented incompletely, but based on standard definitions, rac{4}{3} x is not a valid linear combination in this context. A valid linear combination would look like $Ax+By=C$ where the ratio $A/B$ (if $B eq 0$) or $B/A$ (if $A eq 0$) is consistent with the original system's slopes, or more directly, where it can be expressed as $c_1 imes ( ext{Eq 1}) + c_2 imes ( ext{Eq 2})$. Since rac{4}{3} x lacks a y-term and a constant, it's not a standard linear combination outcome from this system.

Final Answer Structure: The system has no solution because the lines are parallel. A linear combination of these equations will produce another equation representing a line parallel to the original two. This means the coefficients of x and y in the resulting equation will be proportional to the original coefficients, or identically the same if we scale appropriately. The simplified system is $4x+15y=90$ and $4x+15y=12$. Any valid linear combination will take the form $4x+15y=k$ for some constant $k$. The option rac{4}{3} x does not fit this form as it lacks a y-term and a constant, and thus cannot represent a linear combination of the given system in its standard interpretation. Therefore, if rac{4}{3} x is presented as an option, it's the one that could NOT be a linear combination.