Solving Linear Equations: Substitution, Elimination, And Determinants

Hey guys! Ever stumbled upon a set of equations and thought, "Whoa, where do I even begin?" Well, fear not! Today, we're diving deep into the world of linear equations and exploring some awesome methods to crack the code: substitution, elimination, and determinants. We'll tackle a specific problem and see how each technique works its magic. Get ready to flex those math muscles and discover which method reigns supreme! Let's solve this problem and find out which method is the best.

Let's start with the equations you gave us, shall we?

1. 7a + 3b + 2c = 4
2. 4a + 8b + 5c = 6
3. 6a + 4b + 9c = 8

These are three linear equations with three unknowns: a, b, and c. Our goal? To find the values of a, b, and c that make all three equations true. It's like finding the secret combination to unlock a treasure chest! So, let's explore the methods and see how to get the values for a, b, and c.

(a) Unveiling the Secrets: Substitution, Elimination, and Determinants

Alright, buckle up, because we're about to put on our detective hats and solve these linear equations using three different approaches. Each method has its own unique flair, so let's break them down one by one, and in the end, we'll see which method is the best!

(1) Substitution: The Puzzle Solver

Substitution is like a clever jigsaw puzzle. The main idea is to isolate one variable in one equation and then plug its value into the other equations. This reduces the number of variables in the remaining equations, making them easier to solve. Let's start with our first equation: 7a + 3b + 2c = 4. We can solve for 'a' to get a = (4 - 3b - 2c)/7. Next, we would substitute this expression into the other two equations. So let's start with the second equation, 4a + 8b + 5c = 6. By substituting 'a' we will get: 4((4 - 3b - 2c)/7) + 8b + 5c = 6.

Now, simplifying things a bit can get us 16/7 - 12b/7 - 8c/7 + 8b + 5c = 6, and after combining some like terms, this can be written as 44b/7 + 27c/7 = 26/7. We can follow the same process with the third equation, 6a + 4b + 9c = 8. By substituting 'a' we get: 6((4 - 3b - 2c)/7) + 4b + 9c = 8. And if we simplify this, it will become: 24/7 - 18b/7 - 12c/7 + 4b + 9c = 8. Combining terms gives us: 10b/7 + 51c/7 = 32/7. Great, now we are ready to solve this new set of equations with two variables, b and c, using the substitution method again. This process might seem a bit long, but trust me, it's a surefire way to get to the solution! We could solve for 'b' in the equation 44b/7 + 27c/7 = 26/7 to get b = (26 - 27c) / 44, and substituting this into the equation 10b/7 + 51c/7 = 32/7, you'd find a solution for c. Back-substituting these values will help you find b, and then finally, a.

This process, though effective, can be a bit tedious with fractions. But hey, it gets the job done! Just take your time, double-check your calculations, and you'll eventually arrive at the correct values for a, b, and c. This method will take more steps than the others, but it will solve your linear equations. Remember, in this approach, we isolate one variable and substitute its expression into other equations. We repeat this process until we have an equation with only one variable, which we can then solve. Then, we back-substitute the value we found to find the remaining variables.

(2) Elimination: The Sibling Destroyer

Elimination, on the other hand, is like playing a game of math-based whack-a-mole. The idea is to manipulate the equations (multiply them by a constant) so that when you add or subtract them, one of the variables gets "eliminated" (its coefficients cancel out). Let's take the same set of linear equations.

First, we want to eliminate one variable by combining two of the equations. Let's aim to eliminate 'a'. To do this, we need to make the coefficients of 'a' in two equations opposites. Let's multiply the first equation by 4 and the second equation by -7. This will give us: 28a + 12b + 8c = 16 and -28a - 56b - 35c = -42. Adding these two equations will give us -44b - 27c = -26. Great, now we have eliminated 'a' from these two equations.

Now, let's eliminate 'a' from another pair of equations. Let's use the first equation and the third equation. Multiply the first equation by 6 and the third equation by -7, which gives us: 42a + 18b + 12c = 24 and -42a - 28b - 63c = -56. Adding these two equations will give us: -10b - 51c = -32. We can now solve these two equations with b and c. The approach here is to get rid of one variable by adding or subtracting the equations. We do this by making the coefficients of the variable we want to eliminate opposites, which will make the variable disappear!

Once we have two equations with two variables, we can repeat the elimination process to solve for the remaining variables, just like we did in the first step. This method is often preferred because it can be more efficient than substitution, especially when dealing with larger systems of equations. It involves manipulating the equations so that when they are added or subtracted, one of the variables is eliminated. We strategically multiply equations by constants and then add or subtract them to eliminate variables. It's a strategic game of getting rid of variables until we can solve for one and then back-substitute to find the rest. This method will also help you solve your linear equations.

(3) Determinants: The Matrix Magician

Determinants take a slightly different approach. This method uses matrices and determinants to solve the system of equations. For each variable, we calculate the determinant of a matrix formed by specific coefficients and then divide by the determinant of the main coefficient matrix. Let's use the same set of linear equations.

First, we need to set up the coefficient matrix (let's call it D):

D = | 7 3 2 | | 4 8 5 | | 6 4 9 |

We then calculate the determinant of D, which is a number associated with the matrix. To calculate the determinant, you can use the following formula: det(D) = 7(89 - 54) - 3(49 - 56) + 2(44 - 86). Simplifying this will give us the determinant of D. The process continues by replacing one column of the coefficient matrix with the constant terms from the equations and calculating the determinant of the resulting matrix. For example, to find 'a,' we replace the first column of matrix D with the constant terms: 4, 6, and 8. The new matrix, Da, will be:

Da = | 4 3 2 | | 6 8 5 | | 8 4 9 |

Then, we calculate the determinant of Da, which will give us the determinant of a. We repeat this process for b and c, replacing the second and third columns, respectively, with the constant terms and calculating the determinants. Finally, we calculate the value of each variable using these formulas: a = det(Da) / det(D), b = det(Db) / det(D), and c = det(Dc) / det(D). This approach might seem complex, but it can be incredibly efficient for larger systems of equations. Using determinants involves calculating the determinants of matrices formed by the coefficients and constants in the equations. The solution is then found by dividing specific determinants. While it might seem a bit abstract, determinants offer a systematic way to solve the equations. This method will also help you solve the linear equations.

(b) Choosing the Champion: Which Method Reigns Supreme?

So, which method is the best, guys? Well, it depends! Each technique has its pros and cons, but here's a quick breakdown to help you decide:

• Substitution: Great for simpler equations, especially when one variable is already isolated or easy to isolate. It can get tedious with more complex equations.
• Elimination: Often the most efficient, particularly for systems with many equations. It's organized and systematic.
• Determinants: Provides a structured method, great for larger systems. You need to be comfortable with matrix operations and calculations.

In this particular case, elimination is likely to be the most efficient method due to the structure of the equations. However, substitution can also work, but it might involve more algebraic manipulation. Determinants can work well, but the calculation of determinants can be time-consuming, especially by hand. Remember, there's no one-size-fits-all approach! It's all about choosing the tool that best suits the job and your comfort level. The method you use really depends on the complexity of the equations and the specific problem you are facing. Each method will solve the set of linear equations you are presented with.

Conclusion

There you have it, guys! We've journeyed through the realms of linear equations, mastering substitution, elimination, and determinants. Now you have the knowledge and tools to confidently tackle any system of linear equations. So keep practicing, and don't be afraid to experiment with different methods. Happy solving!