Solving System Of Equations: Elimination, Substitution, Determinants

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Hey guys! Let's dive into solving a system of linear equations. We've got two equations here, and our mission is to find the values of x and y that make both equations true. We’re going to tackle this using three awesome methods: elimination, substitution, and determinants. Buckle up, because we're about to unravel this mathematical puzzle!

Understanding the System of Equations

Before we jump into the solving techniques, let's make sure we understand what we're dealing with. We have the following system:

  1. 6x + 4y = 12
  2. 4x + 2y = 8

A system of equations is simply a set of two or more equations that we solve simultaneously. In this case, we have two linear equations, each with two variables (x and y). Our goal is to find the one pair of values for x and y that works for both equations. Think of it like finding the exact spot where two lines intersect on a graph. That intersection point represents the solution that satisfies both equations.

Solving systems of equations is a fundamental concept in algebra and has tons of applications in real life. From figuring out mixtures in chemistry to optimizing business strategies, understanding how to solve these systems is super valuable. Plus, mastering these techniques sets you up for more advanced math down the road. So, let's get started and see how we can crack this code!

Method 1: Elimination - The Art of Wiping Out Variables

Elimination is a clever technique where we manipulate the equations to eliminate one of the variables. The core idea is to make the coefficients (the numbers in front of the variables) of either x or y opposites of each other in the two equations. When we add the equations together, that variable will disappear, leaving us with a single equation in one variable. This is where the magic happens, guys! We make one of the variables go poof!

Here's how it works for our system:

  1. 6x + 4y = 12
  2. 4x + 2y = 8

To eliminate y, let's multiply the second equation by -2. This will give us a -4y term, which is the opposite of the 4y in the first equation:

-2 * (4x + 2y) = -2 * 8

-8x - 4y = -16

Now, we have:

  1. 6x + 4y = 12
  2. -8x - 4y = -16

Next, we add the two equations together. Notice how the y terms cancel each other out:

(6x + 4y) + (-8x - 4y) = 12 + (-16)

6x - 8x = -4

-2x = -4

Now we have a simple equation in just x. We can solve for x by dividing both sides by -2:

x = -4 / -2

x = 2

Awesome! We've found the value of x! Now that we know x, we can substitute this value back into either of the original equations to solve for y. Let's use the first equation:

6x + 4y = 12

Substitute x = 2:

6(2) + 4y = 12

12 + 4y = 12

Subtract 12 from both sides:

4y = 0

Divide by 4:

y = 0

And there we have it! We've found that x = 2 and y = 0. This is the solution to our system using the elimination method. It's like we're mathematical detectives, strategically eliminating clues until we reveal the answer!

Method 2: Substitution - The Art of Swapping Variables

The substitution method is another slick way to solve systems of equations. Instead of eliminating a variable, we solve one equation for one variable and then substitute that expression into the other equation. It's like we're swapping out one variable for an equivalent expression. This turns our two-variable problem into a single-variable problem that's much easier to solve. It's all about strategic swapping, guys!

Let's apply this to our system:

  1. 6x + 4y = 12
  2. 4x + 2y = 8

First, we need to pick one equation and solve it for one variable. The second equation looks a bit simpler, so let's solve it for y:

4x + 2y = 8

Subtract 4x from both sides:

2y = 8 - 4x

Divide both sides by 2:

y = 4 - 2x

Now we have an expression for y in terms of x. Next, we substitute this expression for y into the other equation (the one we didn't use yet), which is the first equation:

6x + 4y = 12

Substitute y = 4 - 2x:

6x + 4(4 - 2x) = 12

Now we have an equation with just x. Let's simplify and solve for x:

6x + 16 - 8x = 12

Combine like terms:

-2x + 16 = 12

Subtract 16 from both sides:

-2x = -4

Divide by -2:

x = 2

Look familiar? We got x = 2 again! Now that we have x, we can substitute it back into the expression we found for y:

y = 4 - 2x

Substitute x = 2:

y = 4 - 2(2)

y = 4 - 4

y = 0

Boom! We found y = 0, just like before. So, using substitution, we've confirmed that the solution to our system is x = 2 and y = 0. It's like a mathematical puzzle where we carefully swap pieces until the picture becomes clear!

Method 3: Determinants - The Matrix Magic

Now, let's tackle the system using determinants, which is a more advanced technique that uses matrices. Don't worry, it's not as scary as it sounds! Determinants give us a systematic way to solve systems of equations, especially when we're dealing with larger systems (more equations and variables). It's like we're using a mathematical machine to crank out the solution, guys!

The method of determinants, also known as Cramer's Rule, involves setting up matrices and calculating their determinants. A matrix is simply a rectangular array of numbers arranged in rows and columns. The determinant is a special number that we can calculate from a square matrix (a matrix with the same number of rows and columns).

For our system:

  1. 6x + 4y = 12
  2. 4x + 2y = 8

We can set up the following matrices:

  • Coefficient Matrix (D): This matrix consists of the coefficients of x and y in our equations.

    D = | 6 4 |

    | 4  2 |

  • X Matrix (Dx): We replace the x coefficients in the coefficient matrix with the constants from the right side of the equations.

    Dx = | 12 4 |

     | 8   2 |

  • Y Matrix (Dy): Similarly, we replace the y coefficients with the constants.

    Dy = | 6 12 |

     | 4  8  |

Now, let's calculate the determinants of these matrices. For a 2x2 matrix | a b |, the determinant is (ad - bc). | c d |

  • Determinant of D (|D|):

    |D| = (6 * 2) - (4 * 4) = 12 - 16 = -4

  • Determinant of Dx (|Dx|):

    |Dx| = (12 * 2) - (4 * 8) = 24 - 32 = -8

  • Determinant of Dy (|Dy|):

    |Dy| = (6 * 8) - (12 * 4) = 48 - 48 = 0

Cramer's Rule tells us that:

x = |Dx| / |D|

y = |Dy| / |D|

So, let's plug in our values:

x = -8 / -4 = 2

y = 0 / -4 = 0

Ta-da! We've arrived at the same solution: x = 2 and y = 0. Determinants might seem a bit more involved, but they provide a powerful and systematic approach to solving systems of equations, especially when things get more complex. It's like using a mathematical Swiss Army knife!

Conclusion: Mission Accomplished!

Alright guys, we've successfully solved our system of equations using three different methods: elimination, substitution, and determinants. We found that the solution is x = 2 and y = 0. Each method gave us the same answer, which is a great way to check our work and build confidence in our solving skills. It’s like we've proven the same theorem in three different ways – mathematical mastery!

Each of these techniques has its own strengths and is useful in different situations. Elimination is great when the coefficients line up nicely for cancellation. Substitution is handy when one equation is easily solved for one variable. Determinants provide a systematic approach that's particularly useful for larger systems. Knowing all three methods gives you a versatile toolkit for tackling any system of equations that comes your way. It’s like having a variety of tools in your mathematical toolbox!

So, keep practicing these methods, and you'll become a system-solving superstar in no time! And remember, math isn't just about getting the right answer; it's about understanding the process and the logic behind it. Keep exploring, keep learning, and keep having fun with math, guys! You've got this!