Solve These Linear Equations

When we're faced with a system of linear equations like the one presented:

$$\begin{array}{l} 3x+5y=10 \\ -3x-5y=12 \end{array}$$

We're looking at two distinct lines on a graph, and our goal is to find the point (or points) where these lines intersect. This intersection point represents the solution that satisfies both equations simultaneously. In this particular case, we have a fascinating scenario that reveals a key concept in algebra. Let's dive into the methods we can use to tackle this system and understand what the outcome tells us. The most common approaches are substitution and elimination. The substitution method involves solving one equation for one variable and then plugging that expression into the other equation. The elimination method, on the other hand, aims to add or subtract the equations in a way that cancels out one of the variables. For our specific system, the elimination method looks particularly promising because of the coefficients. Notice how the coefficients of $x$ (3 and -3) and $y$ (5 and -5) are opposites. This is a strong indicator that adding the two equations together will simplify things considerably.

Let's apply the elimination method. We'll add the first equation to the second equation, term by term:

$$(3x + 5y) + (-3x - 5y) = 10 + 12$$

Combining like terms, we get:

$$(3x - 3x) + (5y - 5y) = 22$$

This simplifies to:

$$0x + 0y = 22$$

Which further reduces to:

$$0 = 22$$

Now, this is where things get interesting. The statement "0 = 22" is false. It's an impossibility; zero can never equal twenty-two. When we arrive at a false statement like this after correctly applying algebraic steps, it means there is no solution to this system of equations. In graphical terms, this signifies that the two lines represented by these equations are parallel. Parallel lines, by definition, never intersect, and therefore, there's no common point (x, y) that lies on both lines. It's crucial to understand that not all systems of equations have a single, unique solution. Some systems, like this one, are designed to illustrate the concept of no solution, while others might have infinitely many solutions (if the two equations represented the exact same line).

Understanding Parallel Lines in Systems of Equations

To truly grasp why this system has no solution, let's explore the concept of parallel lines in the context of linear equations. A linear equation in two variables, $x$ and $y$, typically represents a straight line on a Cartesian plane. The standard form of a linear equation is $Ax + By = C$. When we have a system of two such equations, we are essentially asking where these two lines cross paths. If the lines are parallel, they maintain a constant distance from each other and will never meet. The defining characteristic of parallel lines is that they have the same slope but different y-intercepts. Let's rewrite our given equations in slope-intercept form ($y = mx + b$), where $m$ is the slope and $b$ is the y-intercept.

For the first equation, $3x + 5y = 10$:

Subtract $3x$ from both sides:

$$5y = -3x + 10$$

Divide by 5:

y = -rac{3}{5}x + 2

Here, the slope ($m_1$) is -rac{3}{5} and the y-intercept ($b_1$) is 2.

Now, let's do the same for the second equation, $-3x - 5y = 12$:

Add $3x$ to both sides:

$$-5y = 3x + 12$$

Divide by -5:

y = -rac{3}{5}x - rac{12}{5}

In this second equation, the slope ($m_2$) is also -rac{3}{5}, and the y-intercept ($b_2$) is -rac{12}{5}.

As you can see, both lines have the exact same slope (-rac{3}{5}). This confirms they are parallel. However, their y-intercepts are different (2 and -rac{12}{5}). Because the slopes are identical and the y-intercepts are distinct, these lines will never intersect. This geometrical interpretation perfectly aligns with the algebraic result of $0 = 22$, which indicated no solution. The system is inconsistent. Therefore, when solving systems of linear equations, always be prepared for scenarios that don't yield a single numerical answer; sometimes, the answer is that there is no answer at all.

The Concept of Inconsistent Systems

An inconsistent system of linear equations is one that has no solution. This is precisely what we've encountered with the given equations. Inconsistency arises when the equations in the system contradict each other in such a way that no values for the variables can satisfy all conditions simultaneously. We've already demonstrated this algebraically by arriving at a false statement ($0 = 22$) and geometrically by showing that the lines represented by the equations are parallel and thus never intersect. It's important to distinguish inconsistent systems from consistent systems. Consistent systems are those that have at least one solution. Consistent systems can be further classified into two types: independent and dependent.

An independent consistent system has exactly one unique solution. This is the most common type of system encountered, where the two lines intersect at a single point. For example, if our system was $x + y = 5$ and $x - y = 1$, we could solve it to find $x=3$ and $y=2$. This represents two distinct lines crossing at (3, 2).

A dependent consistent system, on the other hand, has infinitely many solutions. This occurs when the two equations in the system are essentially the same line, just written in a different form. For instance, if we had the system $x + y = 3$ and $2x + 2y = 6$, multiplying the first equation by 2 gives us the second equation. This means both equations describe the exact same line, and every point on that line is a solution to the system. In such cases, when you try to solve algebraically (e.g., using elimination), you'll often end up with a true statement like $0 = 0$. This indicates that the equations are dependent and any point satisfying one will satisfy the other.

Our original problem, with $3x + 5y = 10$ and $-3x - 5y = 12$, is a classic example of an inconsistent system. The contradiction arises because the left-hand sides of the equations are exact opposites ($3x + 5y$ and $-3x - 5y$), implying that $3x + 5y = -(3x + 5y)$. If this were true, then $3x + 5y$ would have to equal 0. However, the right-hand sides of the equations are different ($10$ and $12$). This creates an irreconcilable conflict: the same expression ($3x + 5y$) is claimed to be equal to two different numbers (10 and -12, if we were to manipulate the second equation's left side to match the first). This fundamental contradiction is why no solution exists. Understanding these classifications—inconsistent, independent consistent, and dependent consistent—provides a complete picture of the possible outcomes when solving systems of linear equations.

When Do Systems Have No Solution?

Systems of linear equations have no solution when the equations present contradictory information, making it impossible for any set of variable values to satisfy all conditions simultaneously. We've seen this clearly in our example where adding the two equations led to the false statement $0 = 22$. This outcome is a direct result of the mathematical relationship between the equations. Specifically, a system of two linear equations in two variables will have no solution if, after manipulation, the equations simplify to the form where the variable terms cancel out completely, leaving a false equality. Graphically, this corresponds to the lines being parallel and distinct. Let's consider some general conditions.

Suppose we have a system:

$$A_1x + B_1y = C_1$$

$$A_2x + B_2y = C_2$$

If the ratio of the coefficients of $x$ is equal to the ratio of the coefficients of $y$, but this ratio is not equal to the ratio of the constant terms, then the system has no solution. Mathematically, this can be expressed as:

rac{A_1}{A_2} = rac{B_1}{B_2} eq rac{C_1}{C_2}

Let's check this condition with our given equations: $3x + 5y = 10$ and $-3x - 5y = 12$. Here, $A_1 = 3$, $B_1 = 5$, $C_1 = 10$, and $A_2 = -3$, $B_2 = -5$, $C_2 = 12$.

Now, let's compute the ratios:

rac{A_1}{A_2} = rac{3}{-3} = -1

rac{B_1}{B_2} = rac{5}{-5} = -1

rac{C_1}{C_2} = rac{10}{12} = rac{5}{6}

We can see that rac{A_1}{A_2} = rac{B_1}{B_2} = -1, but -1 eq rac{5}{6}. Thus, the condition rac{A_1}{A_2} = rac{B_1}{B_2} eq rac{C_1}{C_2} is met. This confirms that the system is inconsistent and has no solution. The equality of the ratios of the coefficients ($A_1/A_2 = B_1/B_2$) implies that the lines have the same slope, meaning they are parallel. The inequality of the ratio of the constant terms ($ eq C_1/C_2 $) signifies that they are distinct parallel lines, not the same line. This ratio test is a powerful shortcut for determining the nature of a system's solution without necessarily solving for the variables themselves. It's a fundamental concept in understanding the geometry and algebra of linear systems.

For additional resources on solving systems of linear equations, you can refer to Khan Academy or Paul's Online Math Notes.