Solving Systems Of Equations: Inconsistent Vs. Infinite Solutions

When you're diving into the world of algebra, one of the most common tasks is to solve systems of equations. These systems involve two or more equations with the same variables, and our goal is usually to find the values of these variables that satisfy all equations simultaneously. Today, we're going to tackle a specific system:

• $3x - 4y = 5$
• $-6x + 8y = -10$

We'll not only find the solution but also determine if the system is inconsistent (meaning there's no solution) or has infinitely many solutions. If it's the latter, we'll learn how to express that solution set with one variable arbitrary.

Understanding System Types: A Quick Refresher

Before we get our hands dirty with the numbers, let's quickly review the different types of solutions a system of linear equations can have. Think of these equations as lines on a graph. The solution to the system is where these lines intersect.

1. Unique Solution: This happens when the two lines intersect at exactly one point. In algebraic terms, you'll find specific values for x and y that make both equations true.
2. No Solution (Inconsistent System): This occurs when the two lines are parallel and never intersect. Algebraically, when you try to solve the system, you'll arrive at a contradiction, like $0 = 5$. This impossible statement tells you there's no pair of (x, y) that can satisfy both equations.
3. Infinitely Many Solutions: This is the case when the two equations actually represent the same line. Every point on that line is a solution to both equations. Algebraically, when you try to solve the system, you'll end up with an identity, like $0 = 0$. This true statement indicates that any (x, y) pair that satisfies one equation will automatically satisfy the other.

Our goal with the system $3x - 4y = 5$ and $-6x + 8y = -10$ is to figure out which of these three categories it falls into.

Tackling the System: Step-by-Step Solution

Let's use the method of elimination to solve our system. The idea here is to manipulate one or both equations so that when we add them together, one of the variables cancels out.

Our system is:

Equation 1: $3x - 4y = 5$ Equation 2: $-6x + 8y = -10$

Notice the coefficients of x and y in the second equation. If we multiply the first equation by 2, we get:

$2 * (3x - 4y) = 2 * 5$ $6x - 8y = 10$

Now, let's compare this new equation with our original Equation 2:

New Equation 1: $6x - 8y = 10$ Equation 2: $-6x + 8y = -10$

Let's add these two equations together:

$(6x - 8y) + (-6x + 8y) = 10 + (-10)$ $6x - 6x - 8y + 8y = 10 - 10$ $0x + 0y = 0$ $0 = 0$

Analyzing the Result: What Does $0 = 0$ Mean?

We've arrived at the statement $0 = 0$. As we discussed earlier, this is a true statement. In the context of solving systems of equations, a true statement like this indicates that the two original equations are essentially the same line, just written in a different form. This means that any point that lies on the first line also lies on the second line, and vice versa. Therefore, there isn't just one solution; there are infinitely many solutions.

Writing the Solution Set with an Arbitrary Variable

When a system has infinitely many solutions, we can't list every single (x, y) pair. Instead, we express the solution set in terms of one variable being arbitrary. This means we pick one variable (usually y, but sometimes x is easier) and let it represent any real number. Then, we write the other variable in terms of this arbitrary one.

Let's go back to our original Equation 1: $3x - 4y = 5$.

We want to express x in terms of y. First, let's isolate the x term:

$3x = 5 + 4y$

Now, divide by 3 to solve for x:

x = rac{5 + 4y}{3}

So, our solution set can be written as: ( rac{5 + 4y}{3}, y ), where y can be any real number.

Alternatively, if we wanted to express y in terms of x, we would start again with $3x - 4y = 5$:

$-4y = 5 - 3x$

Divide by -4:

y = rac{5 - 3x}{-4}

y = rac{3x - 5}{4}

In this case, the solution set would be (x, rac{3x - 5}{4}), where x can be any real number.

Both forms are correct ways to represent the infinitely many solutions. The prompt specifically asked to write the solution set with y arbitrary, so we'll stick with the first form: ( rac{5 + 4y}{3}, y ).

Conclusion: Infinite Possibilities!

So, to recap, our system of equations:

• $3x - 4y = 5$
• $-6x + 8y = -10$

is a case of infinitely many solutions. This happened because the second equation is simply the first equation multiplied by -2, meaning they represent the exact same line. The mathematical journey led us to the identity $0=0$, a clear sign of this scenario. We successfully expressed the solution set as ( rac{5 + 4y}{3}, y ), where y can be any real number. This means there's a solution for every possible value you choose for y!

Understanding the different types of solutions – unique, inconsistent, or infinite – is a fundamental skill in algebra. It helps us interpret the relationships between equations and provides a complete picture of their interactions. Keep practicing, and you'll become a pro at navigating these systems!

For further exploration into solving systems of linear equations and understanding their graphical interpretations, you can visit Khan Academy's Algebra section or Math is Fun's explanation of systems of equations.