Solve Systems Of Equations: Find The Solution (2,0)

Hey there, math enthusiasts! Ever wondered how to pinpoint the exact solution to a system of equations? It's like being a detective, where each equation is a clue, and the solution is the "aha!" moment. Today, we're going to tackle a classic problem: finding which system of equations has (2,0) as its solution. This means we're looking for a pair of equations where, when you plug in x=2 and y=0, both equations ring true. Let's dive in and see how we can crack this code!

Understanding Systems of Equations and Solutions

Before we get our hands dirty with the specific options, let's quickly recap what a system of equations actually is. In simple terms, it's a collection of two or more equations that share the same set of variables. When we talk about the solution to a system of equations, we're referring to the specific values of those variables that satisfy all the equations in the system simultaneously. For a system with two variables (like x and y), the solution is typically represented as an ordered pair (x, y). In our case, we are given a potential solution: the ordered pair (2, 0). This means we need to find a set of two equations where x = 2 and y = 0 makes both equations correct. Think of it as a lock and key; the solution (2,0) must fit perfectly into both equation "locks" to be valid.

Why is the Solution Unique?

The beauty of a system of equations is that, under certain conditions, it has a unique solution. This uniqueness often arises from the relationships between the lines represented by the equations. If the lines have different slopes, they will intersect at exactly one point – that point is the unique solution. If the lines are parallel and distinct, there's no solution. If the lines are identical, there are infinitely many solutions. Our goal here is to find the system where the intersection point is precisely at (2, 0). This makes our task a process of verification. We don't need to solve from scratch; we just need to test the given point against each system.

Method of Verification: Plugging in the Values

The most straightforward way to determine which system has (2,0) as its solution is by substitution. We'll take the ordered pair (2, 0) – meaning x=2 and y=0 – and plug these values into each equation within each system. If both equations in a given system hold true after substitution, then that system is our winner!

Let's break down how we'll do this for each option. For any given equation, say $Ax + By = C$, we'll substitute x=2 and y=0. The equation becomes $A(2) + B(0) = C$. We then simplify this to $2A = C$. If this statement is true, the point (2,0) satisfies that specific equation. We must repeat this check for the second equation in the system. Only when (2,0) satisfies both equations in a pair do we have our solution.

Step-by-Step Substitution Process

We'll go through each option systematically:

• Option A:

  • Equation 1: $6x - 4y = 6$
  • Equation 2: $x + 4y = 2$

• Option B:

  • Equation 1: $6x - 4y = 6$
  • Equation 2: $2x + 8y = 2$

• Option C:

  • Equation 1: $x + 4y = 2$
  • Equation 2: $7x = 14$

• Option D:

  • Equation 1: $6x - 4y = 6$
  • Equation 2: $7x = 14$

By performing these checks, we can confidently identify the correct system. This method is robust and efficient for this type of problem.

Evaluating Option A

Let's start with Option A. Our potential solution is (2, 0), so x=2 and y=0.

• Equation 1: $6x - 4y = 6$ Substitute x=2 and y=0: $6(2) - 4(0) = 6$ $12 - 0 = 6$ $12 = 6$ This statement is false. Since the first equation is not satisfied, Option A cannot be the correct system, regardless of whether the second equation works.

This initial check is crucial. If even one equation in the system fails the test, the entire system is disqualified. It saves us time and effort from checking further when the answer is already apparent.

Evaluating Option B

Now, let's move on to Option B. Again, our test point is (2, 0).

• Equation 1: $6x - 4y = 6$ Substitute x=2 and y=0: $6(2) - 4(0) = 6$ $12 - 0 = 6$ $12 = 6$ This statement is false. Similar to Option A, the first equation in Option B does not hold true for the point (2, 0). Therefore, Option B is also not the correct system.

It's interesting to note that both Option A and Option B share the same first equation, and this equation is the one causing the problem. This highlights how crucial it is for every equation in the system to be satisfied by the proposed solution. If one fails, the whole system fails.

Evaluating Option C

Let's examine Option C. Our point is (2, 0).

• Equation 1: $x + 4y = 2$ Substitute x=2 and y=0: $(2) + 4(0) = 2$ $2 + 0 = 2$ $2 = 2$ This statement is true. So far, so good! The first equation is satisfied.

• Equation 2: $7x = 14$ Substitute x=2: $7(2) = 14$ $14 = 14$ This statement is also true. Since both equations in Option C are satisfied by the point (2, 0), this is our likely candidate for the correct system.

This is where we get excited! When both equations check out, we've found the system that precisely contains the given solution. It's a moment of triumph in our mathematical detective work.

Evaluating Option D

Finally, let's check Option D to be absolutely sure. Our point is (2, 0).

• Equation 1: $6x - 4y = 6$ Substitute x=2 and y=0: $6(2) - 4(0) = 6$ $12 - 0 = 6$ $12 = 6$ This statement is false. As we saw in Options A and B, this first equation is problematic for the point (2,0). Since the first equation is not satisfied, Option D cannot be the correct system.

It's good practice to check all options, especially if there's any doubt or if you're in a testing scenario. In this case, our previous findings strongly pointed to Option C, and checking Option D confirms that it also fails due to the same initial equation.

Conclusion: The Solution Revealed!

After meticulously plugging the values x=2 and y=0 into each equation of every system, we've reached a definitive conclusion. Option C is the only system where both equations are simultaneously true for the point (2, 0).

Let's recap why:

• Option A: Failed Equation 1 ($12 eq 6$).
• Option B: Failed Equation 1 ($12 eq 6$).
• Option C: Satisfied Equation 1 ($2 = 2$) AND Equation 2 ($14 = 14$).
• Option D: Failed Equation 1 ($12 eq 6$).

Therefore, the system of equations that has (2,0) as its solution is:

$x + 4y = 2$ $7x = 14$

Keep practicing these substitution techniques, and you'll become a pro at solving systems of equations in no time! It's all about methodical checking and understanding what a solution truly represents.

For further exploration into systems of equations and their graphical interpretations, you might find the resources at Khan Academy very helpful. They offer a wealth of information and practice problems that can solidify your understanding.