Infinite Solutions: Find 'c' Value In Linear Equations

Hey guys! Let's dive into a cool math problem today that involves finding the value of a constant in a system of linear equations, given that the system has infinitely many solutions. This is a classic problem that combines algebra and a bit of insightful thinking about what it means for equations to have infinite solutions. So, buckle up, and let's get started!

Understanding the Problem

First, let's break down the problem. We're given a system of two linear equations:

1. kx - 10y = c
2. (1/2)x - (1/3)y = 2

We need to find the value of c such that this system has infinitely many solutions. Remember, a system of linear equations can have one solution, no solution, or infinitely many solutions. The key here is the condition for infinite solutions.

Infinitely many solutions occur when the two equations represent the same line. This means one equation is just a multiple of the other. If you graph these equations, they would overlap perfectly. This gives us a crucial strategy for solving the problem: we need to make the two equations proportional to each other.

To achieve infinitely many solutions, the ratios of the coefficients of x, y, and the constant terms must be equal*. This concept is vital. Let's denote the coefficients and constants from the equations as follows:

• Equation 1: a₁ = k, b₁ = -10, c₁ = c
• Equation 2: a₂ = 1/2, b₂ = -1/3, c₂ = 2

For infinitely many solutions, we must have:

a₁ / a₂ = b₁ / b₂ = c₁ / c₂

This is our roadmap. Now, let's apply this understanding to solve for c.

Solving for k

Before we can find c, we'll need to determine the value of k. We can use the ratios of the coefficients of x and y to find k. So, from our condition for infinitely many solutions, we have:

k / (1/2) = (-10) / (-1/3)

Let's simplify this equation. Dividing by a fraction is the same as multiplying by its reciprocal, so we can rewrite the equation as:

2k = (-10) * (-3)

2k = 30

Now, divide both sides by 2:

k = 15

Great! We've found that k = 15. This is a significant step because it allows us to relate the first equation more directly to the second. Knowing k, we can now move on to finding c.

Finding the Value of c

Now that we know k, we can use the ratios involving the constant terms to solve for c. We have:

k / (1/2) = c / 2

Since we found that k = 15, we can substitute that into the equation:

15 / (1/2) = c / 2

Simplify the left side by multiplying 15 by the reciprocal of 1/2:

30 = c / 2

To solve for c, multiply both sides by 2:

c = 30 * 2

c = 60

And there we have it! The value of c that makes the system have infinitely many solutions is 60. This means that if we substitute c = 60 and k = 15 into the original system of equations, the two equations will represent the same line.

Verification

To make sure we're on the right track, let's verify our solution. Substitute k = 15 and c = 60 back into the first equation:

15x - 10y = 60

Now, let's simplify this equation by dividing all terms by 5:

3x - 2y = 12

Next, let’s manipulate the second original equation to look similar:

(1/2)x - (1/3)y = 2

To eliminate fractions, we multiply the entire equation by 6 (the least common multiple of 2 and 3):

6 * [(1/2)x - (1/3)y] = 6 * 2

3x - 2y = 12

Notice that the simplified forms of both equations are identical: 3x - 2y = 12. This confirms that the two equations represent the same line when k = 15 and c = 60, and thus, the system has infinitely many solutions. This step is crucial for ensuring the accuracy of our solution.

Alternative Approach: Scalar Multiplication

Another way to think about this problem is through scalar multiplication. If two linear equations have infinitely many solutions, it means one equation is a scalar multiple of the other. In other words, you can multiply one of the equations by a constant to get the other equation.

Let's look at our original equations again:

1. kx - 10y = c
2. (1/2)x - (1/3)y = 2

We want to find a scalar, let's call it s, such that when we multiply the second equation by s, we get the first equation:

s * [(1/2)x - (1/3)y] = kx - 10y = c

Distributing s across the second equation, we get:

(s/2)x - (s/3)y = kx - 10y = c

Now we can equate the coefficients:

• s/2 = k
• -s/3 = -10
• c = 2s

From the second equation, -s/3 = -10, we can solve for s:

• s = 30

Now that we have s, we can find k and c:

• k = s/2 = 30/2 = 15
• c = 2s = 2 * 30 = 60

This approach gives us the same values for k and c as before, further validating our solution.

Common Pitfalls to Avoid

When solving problems like this, there are a few common mistakes students often make. It's good to be aware of these so you can avoid them!

1. Incorrectly setting up the ratios: The ratios of the coefficients and constants must be set up correctly. Make sure you're comparing corresponding terms from the two equations. A mix-up here can lead to completely wrong answers.
2. Forgetting to check all ratios: For infinitely many solutions, all the ratios (a₁ / a₂, b₁ / b₂, and c₁ / c₂) must be equal. Don't just find one pair of equal ratios and assume you're done. Always verify that all ratios match.
3. Arithmetic errors: Simple arithmetic mistakes can throw off your entire solution. Double-check your calculations, especially when dealing with fractions and negative signs.
4. Not verifying the solution: It's always a good idea to plug your values back into the original equations to make sure they work. This is especially important in problems where multiple steps are involved, as it helps catch any errors made along the way.

Real-World Applications

While this problem might seem purely mathematical, systems of linear equations with infinitely many solutions (or no solutions) pop up in various real-world scenarios. They're particularly relevant in fields like economics, engineering, and computer science.

For instance, in economics, you might encounter systems of equations when modeling supply and demand. If the equations are dependent (one is a multiple of the other), you have infinitely many combinations of price and quantity that satisfy the conditions, indicating a need for additional constraints or information to find a unique solution.

In engineering, these systems can arise in circuit analysis or structural analysis. If a system has infinitely many solutions, it could indicate that the circuit or structure is underdetermined, meaning there aren't enough constraints to define a unique state. This can be crucial for ensuring stability and safety.

In computer graphics and linear programming, understanding systems of equations is essential for transformations, optimizations, and constraint satisfaction problems. Knowing when a system has infinite solutions helps in designing efficient algorithms and models.

Conclusion

So, guys, we've successfully navigated this problem and found that for the given system of equations to have infinitely many solutions, the value of c must be 60. We also found that k must be 15. Remember, the key to solving this type of problem is understanding the condition for infinitely many solutions: the equations must represent the same line, meaning their coefficients and constants are proportional. By setting up the correct ratios and solving for the unknowns, we cracked the case!

We also explored an alternative approach using scalar multiplication, which reinforced our understanding and provided another way to verify our results. Plus, we discussed common pitfalls to avoid and highlighted some real-world applications of these concepts. Keep practicing, and you'll become a master at solving these types of problems!