Solving Systems Of Linear Equations

Hey guys! Ever stared at a jumble of equations and wondered how to untangle them? Well, you're in the right place. Today, we're diving deep into the awesome world of solving systems of linear equations, specifically tackling this beast:

$\begin{array}{l} -5 x-6 y+z=11 \\ 3 x+5 y+5 z=-1 \\ -4 x-y+3 z=-2 \end{array}$

This isn't just about crunching numbers; it's about uncovering the hidden truths within these mathematical relationships. Think of each equation as a clue, and when you solve the system, you're piecing together the whole picture. We'll break down the methods, explore the why's, and get you feeling super confident about wrangling these algebraic challenges. So, grab your thinking caps, and let's get this math party started!

Understanding the Beast: What are Systems of Linear Equations?

Alright, so what exactly is a system of linear equations, and why should you even care? Simply put, it's a collection of two or more linear equations that share the same set of variables. In our case, we've got three equations with three variables: $x$, $y$, and $z$. When we talk about solving a system, we're on a mission to find the specific values of these variables that make all the equations true simultaneously. It's like finding the exact spot where multiple paths cross – a unique solution. For linear equations, these paths are straight lines (in 2D) or planes (in 3D and beyond). The solutions are the points where these lines or planes intersect. Sometimes they intersect at a single point (one unique solution), sometimes they overlap along a line or a plane (infinite solutions), and sometimes they never meet (no solution). The goal is to identify that intersection point, or determine if one even exists. Understanding this concept is super crucial because systems of linear equations pop up everywhere. From figuring out the best way to allocate resources in a business to modeling complex physical phenomena in science, and even in computer graphics for rendering 3D objects, these systems are the backbone of many real-world applications. So, when we're presented with a system like the one above, we're not just doing an academic exercise; we're practicing a fundamental problem-solving technique that has broad applicability. The challenge lies in the fact that we have multiple unknowns and multiple constraints, and we need to find the conditions that satisfy all of them. This often involves a methodical approach, like using elimination or substitution, to systematically reduce the complexity of the problem until we isolate the values of our variables. It's a bit like solving a puzzle, where each piece (equation) gives you information to help you find the final picture (the solution). So, let's get ready to roll up our sleeves and dive into the actual methods that will help us conquer this system.

Method 1: The Elimination Method - Making Variables Vanish!

Okay, team, let's talk about one of the most powerful tools in our arsenal for solving systems of linear equations: the elimination method. The core idea here is genius in its simplicity: we strategically add or subtract the equations to make one of the variables disappear – poof! This reduces our three-equation system into a simpler two-equation system, which is way easier to handle. For our specific system:

$\begin{array}{l} -5 x-6 y+z=11 ...(1) \\ 3 x+5 y+5 z=-1 ...(2) \\ -4 x-y+3 z=-2 ...(3) \end{array}$

Our mission is to make either $x$, $y$, or $z$ vanish. Let's pick $z$ to eliminate first. Notice that equation (1) has $+z$ and equation (3) has $+3z$. If we multiply equation (1) by -3, we get $-3z$, which will cancel out nicely when added to equation (3). So, let's do that:

$-3 \times (-5x - 6y + z = 11) \implies 15x + 18y - 3z = -33 \quad ...(1')$

Now, let's add this modified equation (1') to equation (3):

$(15x + 18y - 3z) + (-4x - y + 3z) = -33 + (-2)$

Combine like terms:

$(15x - 4x) + (18y - y) + (-3z + 3z) = -33 - 2$

$11x + 17y + 0z = -35$

$11x + 17y = -35 \quad ...(4)$

Awesome! We've successfully eliminated $z$ and created a new equation (4) with only $x$ and $y$. But we're not done yet. We need another equation with just $x$ and $y$ to solve for them. Let's eliminate $z$ again, but this time using equations (1) and (2). Equation (1) has $+z$ and equation (2) has $+5z$. We can multiply equation (1) by -5 to get $-5z$, which will cancel with the $+5z$ in equation (2).

$-5 \times (-5x - 6y + z = 11) \implies 25x + 30y - 5z = -55 \quad ...(1'')$

Now, add this modified equation (1'') to equation (2):

$(25x + 30y - 5z) + (3x + 5y + 5z) = -55 + (-1)$

Combine like terms:

$(25x + 3x) + (30y + 5y) + (-5z + 5z) = -55 - 1$

$28x + 35y + 0z = -56$

$28x + 35y = -56 \quad ...(5)$

Look at that! We now have a system of two linear equations with two variables:

$\begin{array}{l} 11x + 17y = -35 ...(4) \\ 28x + 35y = -56 ...(5) \end{array}$

We can simplify equation (5) by dividing everything by 7:

$4x + 5y = -8 \quad ...(5')$

Now our system is:

$\begin{array}{l} 11x + 17y = -35 ...(4) \\ 4x + 5y = -8 ...(5') \end{array}$

We can use elimination again on this smaller system to solve for $x$ or $y$. Let's eliminate $y$. We can multiply equation (4) by 5 and equation (5') by -17. This will give us $+85y$ and $-85y$.

$5 \times (11x + 17y = -35) \implies 55x + 85y = -175 \quad ...(4')$

$-17 \times (4x + 5y = -8) \implies -68x - 85y = 136 \quad ...(5'')$

Now, add (4') and (5''):

$(55x + 85y) + (-68x - 85y) = -175 + 136$

$(55x - 68x) + (85y - 85y) = -39$

$-13x = -39$

Divide by -13:

$x = 3$

Boom! We found $x = 3$. The elimination method really shines when you systematically reduce the problem size, making it much more manageable. It's all about being strategic with your multiplications and additions/subtractions to make those pesky variables disappear.

Method 2: The Substitution Method - Swapping Variables Like a Pro

Another super useful technique for solving systems of linear equations is the substitution method. This approach involves solving one of the equations for one variable in terms of the others, and then substituting that expression into the other equations. It's like playing a game of algebraic peek-a-boo, where you reveal one variable's value and then plug it in everywhere else. Let's use our original system again:

$\begin{array}{l} -5 x-6 y+z=11 ...(1) \\ 3 x+5 y+5 z=-1 ...(2) \\ -4 x-y+3 z=-2 ...(3) \end{array}$

The easiest equation to work with for substitution is usually the one where a variable has a coefficient of 1 or -1, because it's simple to isolate. Looking at our system, equation (1) has $+z$ and equation (3) has $-y$. Let's pick equation (3) and solve for $y$ because it has the coefficient -1:

$-4x - y + 3z = -2$

Add $y$ to both sides:

$-4x + 3z = -2 + y$

Add $4x$ and subtract $3z$ from both sides:

$y = 4x - 3z + 2 ...(3')$

Now that we have an expression for $y$ in terms of $x$ and $z$, we're going to substitute this expression into equations (1) and (2). Let's start with equation (1):

$-5x - 6y + z = 11$

Substitute $(4x - 3z + 2)$ for $y$:

$-5x - 6(4x - 3z + 2) + z = 11$

Distribute the -6:

$-5x - 24x + 18z - 12 + z = 11$

Combine like terms:

$(-5x - 24x) + (18z + z) - 12 = 11$

$-29x + 19z - 12 = 11$

Add 12 to both sides:

$-29x + 19z = 23 ...(4)$

Great! We now have one equation (4) with just $x$ and $z$. Now, let's substitute our expression for $y$ into equation (2):

$3x + 5y + 5z = -1$

Substitute $(4x - 3z + 2)$ for $y$:

$3x + 5(4x - 3z + 2) + 5z = -1$

Distribute the 5:

$3x + 20x - 15z + 10 + 5z = -1$

Combine like terms:

$(3x + 20x) + (-15z + 5z) + 10 = -1$

$23x - 10z + 10 = -1$

Subtract 10 from both sides:

$23x - 10z = -11 ...(5)$

Now we have a new system of two equations with two variables:

$\begin{array}{l} -29x + 19z = 23 ...(4) \\ 23x - 10z = -11 ...(5) \end{array}$

We can solve this new system using either elimination or substitution again! Let's use substitution one more time. From equation (5), let's solve for $x$:

$23x = 10z - 11$

$x = \frac{10z - 11}{23} \quad ...(5')$

Now substitute this expression for $x$ into equation (4):

$-29\left(\frac{10z - 11}{23}\right) + 19z = 23$

Multiply both sides by 23 to clear the fraction:

$-29(10z - 11) + 23(19z) = 23(23)$

$-290z + 319 + 437z = 529$

Combine $z$ terms:

$(-290z + 437z) + 319 = 529$

$147z + 319 = 529$

Subtract 319 from both sides:

$147z = 529 - 319$

$147z = 210$

Divide by 147:

$z = \frac{210}{147}$

We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 21:

$z = \frac{10}{7}$

So, $z = \frac{10}{7}$. Now we need to find $x$ and $y$. Let's substitute $z = \frac{10}{7}$ back into our expression for $x$ (equation 5'):

$x = \frac{10(\frac{10}{7}) - 11}{23}$

$x = \frac{\frac{100}{7} - \frac{77}{7}}{23}$

$x = \frac{\frac{23}{7}}{23}$

$x = \frac{23}{7} \times \frac{1}{23}$

$x = \frac{1}{7}$

Finally, we substitute the values of $x = \frac{1}{7}$ and $z = \frac{10}{7}$ back into our expression for $y$ (equation 3'):

$y = 4x - 3z + 2$

$y = 4(\frac{1}{7}) - 3(\frac{10}{7}) + 2$

$y = \frac{4}{7} - \frac{30}{7} + \frac{14}{7}$

$y = \frac{4 - 30 + 14}{7}$

$y = \frac{-12}{7}$

So, with the substitution method, we found $x = \frac{1}{7}$, $y = \frac{-12}{7}$, and $z = \frac{10}{7}$. The substitution method is great when you can easily isolate a variable, and it systematically breaks down the problem.

Verifying Our Solution: Did We Nail It?

Alright guys, we've done the heavy lifting and found a potential solution using both elimination and substitution. But how do we know for sure we didn't mess up along the way? Easy peasy: verification! We take our proposed solution $(x, y, z) = (3, -1, -2)$ from the elimination method (oops, spotted an error in the earlier calculation, let's re-verify with the correct values from substitution, $(x, y, z) = (1/7, -12/7, 10/7)$) and plug it back into each of the original equations. If all three equations hold true, then we've got ourselves a winner!

Let's use the solution from the substitution method: $x = \frac{1}{7}$, $y = \frac{-12}{7}$, $z = \frac{10}{7}$.

Equation (1): $-5x - 6y + z = 11$

$-5(\frac{1}{7}) - 6(\frac{-12}{7}) + \frac{10}{7} = \frac{-5}{7} - \frac{-72}{7} + \frac{10}{7}$

$= \frac{-5 + 72 + 10}{7} = \frac{77}{7} = 11$

Success! Equation (1) checks out.

Equation (2): $3x + 5y + 5z = -1$

$3(\frac{1}{7}) + 5(\frac{-12}{7}) + 5(\frac{10}{7}) = \frac{3}{7} + \frac{-60}{7} + \frac{50}{7}$

$= \frac{3 - 60 + 50}{7} = \frac{-7}{7} = -1$

Double Success! Equation (2) is also correct.

Equation (3): $-4x - y + 3z = -2$

$-4(\frac{1}{7}) - (\frac{-12}{7}) + 3(\frac{10}{7}) = \frac{-4}{7} - \frac{-12}{7} + \frac{30}{7}$

$= \frac{-4 + 12 + 30}{7} = \frac{38}{7}$

Uh oh! Equation (3) did not result in -2. This means there was an arithmetic error in our calculations. Let's re-trace the elimination method steps, as it is often cleaner for systems like this. My apologies, guys!

Re-Verification with Elimination Method's Calculated Values

Let's re-examine the elimination method and ensure our calculations were spot on. The process was:

1. Eliminate $z$ from (1) and (3) to get $11x + 17y = -35$ (Eq 4).
2. Eliminate $z$ from (1) and (2) to get $28x + 35y = -56$, which simplifies to $4x + 5y = -8$ (Eq 5').
3. Solve the system: $11x + 17y = -35$ (Eq 4) $4x + 5y = -8$ (Eq 5') Multiply (4) by 5 and (5') by -17: $55x + 85y = -175$ $-68x - 85y = 136$ Adding these gives: $-13x = -39 => x = 3$.

Now, substitute $x = 3$ into equation (5'):

$4(3) + 5y = -8$

$12 + 5y = -8$

$5y = -8 - 12$

$5y = -20$

$y = -4$

Now, substitute $x = 3$ and $y = -4$ into equation (1) to find $z$:

$-5(3) - 6(-4) + z = 11$

$-15 + 24 + z = 11$

$9 + z = 11$

$z = 11 - 9$

$z = 2$

So, the solution from the elimination method is $(x, y, z) = (3, -4, 2)$. Let's verify this solution!

Equation (1): $-5x - 6y + z = 11$

$-5(3) - 6(-4) + 2 = -15 + 24 + 2 = 9 + 2 = 11$

Nailed it! Equation (1) is correct.

Equation (2): $3x + 5y + 5z = -1$

$3(3) + 5(-4) + 5(2) = 9 - 20 + 10 = -11 + 10 = -1$

Still on a roll! Equation (2) is correct.

Equation (3): $-4x - y + 3z = -2$

$-4(3) - (-4) + 3(2) = -12 + 4 + 6 = -8 + 6 = -2$

Perfect! All three equations hold true with the solution $(x, y, z) = (3, -4, 2)$. It's incredibly important to double-check your work, especially when dealing with multiple steps and negative numbers. My apologies for the initial mix-up, but this is a great learning moment about the importance of verification!

The Beauty of Matrix Methods: Cramer's Rule & Gaussian Elimination

While elimination and substitution are fantastic for smaller systems, as things get bigger, we often turn to more systematic methods, especially those that can be programmed into computers. Two heavy hitters here are Cramer's Rule and Gaussian Elimination (which is the basis for the elimination method we used but formalized using matrices).

Cramer's Rule: Determinants to the Rescue!

Cramer's Rule is a slick method that uses determinants to find the solution for each variable. A determinant is a special scalar value that can be calculated from a square matrix. For a 3x3 system like ours:

$\begin{array}{l} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{array}$

The determinant of the coefficient matrix, often denoted as $D$, is:

$D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

To find $x$, we replace the first column (the coefficients of $x$) with the constants ($d_1, d_2, d_3$) to get $D_x$, and then $x = \frac{D_x}{D}$. Similarly, for $y$ and $z$:

$D_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix} => y = \frac{D_y}{D}$

$D_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix} => z = \frac{D_z}{D}$

Calculating determinants for 3x3 matrices can be a bit tedious by hand (using Sarrus's rule or cofactor expansion), but it's a very direct computational approach. If $D=0$, the system either has no solution or infinite solutions, and Cramer's Rule can't be directly applied. Our system is:

$\begin{array}{l} -5 x-6 y+z=11 \\ 3 x+5 y+5 z=-1 \\ -4 x-y+3 z=-2 \end{array}$

So, $a_1=-5, b_1=-6, c_1=1, d_1=11$, etc. Calculating these determinants would eventually lead to the same solution $(3, -4, 2)$, provided $D \neq 0$.

Gaussian Elimination: Row Operations to Simplicity!

Gaussian elimination is essentially a formalized version of the elimination method using matrices. We represent the system as an augmented matrix:

$[-5 -6 1 | 11]$ $[ 3 5 5 | -1]$ $[-4 -1 3 | -2]$

The goal is to use elementary row operations (swapping rows, multiplying a row by a non-zero scalar, adding a multiple of one row to another) to transform the left side of the matrix into an identity matrix (1s on the diagonal, 0s elsewhere). The corresponding right side will then be the solution vector. This process systematically zeros out elements below the diagonal and then above the diagonal (Jordan elimination), making the solution clear. It's a robust method, especially for larger systems or when dealing with potential inconsistencies (no solution) or dependencies (infinite solutions), which manifest as rows of zeros.

Conclusion: Mastering the Math Maze

So there you have it, folks! We've navigated the twists and turns of solving systems of linear equations. We tackled our specific problem using the elimination method, which proved quite effective, and explored the substitution method as an alternative. We also touched upon the elegance of matrix methods like Cramer's Rule and Gaussian Elimination, which are invaluable for more complex scenarios and computational approaches. The key takeaway is that while the methods might seem daunting at first, they are all systematic processes designed to break down complex problems into manageable steps. Practice is your best friend here. The more systems you solve, the more comfortable you'll become with the calculations, spotting potential errors, and choosing the most efficient method for a given problem. Remember that verification step – it's your safety net! Don't be afraid to re-trace your steps if something doesn't add up. Keep practicing, keep questioning, and you'll become a master at solving these mathematical mazes. Happy solving!