Solving A System Of 3 Linear Equations (SPLTV)

Hey guys! Today, we're diving into solving a system of three-variable linear equations, often called SPLTV. This might sound intimidating, but trust me, with a step-by-step approach, it's totally manageable. We've got three equations:

1. $5y - 8z = -19$
2. $5x - 8z = 6$
3. $3x - 2y = 12$

Let's break down how to solve this bad boy!

Step 1: Choose a Variable to Eliminate

The key to solving SPLTV systems is to eliminate one variable at a time. Looking at our equations, notice that 'z' appears in the first two equations. It might be easiest to start by eliminating 'z'. To do this, we can manipulate the first two equations so that the coefficients of 'z' are opposites, then add the equations together. Alternatively, we can eliminate either x or y, it only depends on which one you prefer. If we choose to eliminate $z$, we will continue to the next step.

When diving into solving systems of equations, especially SPLTV (Sistem Persamaan Linear Tiga Variabel) which translates to System of Linear Equations with Three Variables, the initial and perhaps most crucial step involves selecting a variable for elimination. This decision isn't arbitrary; it's about optimizing the solution process. The aim is to identify a variable that, when eliminated, simplifies the system and reduces the computational burden. In our case, we're presented with three equations: $5y - 8z = -19$, $5x - 8z = 6$, and $3x - 2y = 12$. A closer look reveals that the variable 'z' appears prominently in the first two equations. This observation makes 'z' a prime candidate for elimination, mainly because its presence in multiple equations provides a direct pathway for manipulation and subsequent removal. To elaborate, the strategy for eliminating 'z' revolves around modifying the equations such that the coefficients of 'z' become opposites. This can be achieved through multiplication by appropriate constants. Once the coefficients are opposites, adding the equations will effectively cancel out 'z', leading to a new equation with only two variables. This new equation, combined with the third original equation (which already contains only 'x' and 'y'), forms a simpler two-variable system that is much easier to solve. However, it's worth noting that the choice of 'z' is not the only viable option. One could also opt to eliminate 'x' or 'y' instead. The decision often depends on personal preference or which manipulation appears easiest. For instance, if eliminating 'x' seems more straightforward due to simpler coefficients, one might choose that route. Similarly, if 'y' appears in a way that allows for easy isolation or elimination, that could be the preferred approach. Ultimately, the goal is to select the variable that minimizes complexity and streamlines the solution process. By making a judicious choice at this initial step, you set the stage for a more efficient and less error-prone solution.

Step 2: Manipulate Equations to Eliminate 'z'

We'll multiply the first equation by 5 and the second equation by -5:

• Equation 1 (multiplied by 5):* $25y - 40z = -95$
• Equation 2 (multiplied by -5):* $-25x + 40z = -30$

Now, add these two equations:

$(25y - 40z) + (-25x + 40z) = -95 + (-30)$

This simplifies to:

$-25x + 25y = -125$

Divide the entire equation by -25:

$x - y = 5$ Let's call this Equation 4.

When manipulating equations to eliminate a chosen variable, in this case 'z', the essence lies in strategically modifying the equations such that the coefficients of the target variable become opposites. This is a critical step that sets the stage for the subsequent elimination and simplification of the system. To achieve this, one employs multiplication, a fundamental algebraic operation that allows us to scale equations without altering their inherent relationships. In our specific scenario, we began with the equations $5y - 8z = -19$ and $5x - 8z = 6$. The goal was to transform these equations so that the 'z' terms would cancel each other out upon addition. To this end, we multiplied the first equation by 5 and the second equation by -5. This yielded two new equations: $25y - 40z = -95$ and $-25x + 40z = -30$. Notice how the 'z' terms now have coefficients that are opposites: -40z and +40z. The next step involves adding these two modified equations together. This is where the magic happens. When we add the left-hand sides and the right-hand sides separately, we get $(25y - 40z) + (-25x + 40z) = -95 + (-30)$. As anticipated, the 'z' terms cancel each other out, leaving us with an equation that involves only 'x' and 'y': $-25x + 25y = -125$. This equation represents a significant simplification of the original system. However, it's not the end of the road. To further refine this equation and make it more manageable, we can divide the entire equation by a common factor. In this case, we divided by -25, resulting in $x - y = 5$. This final equation, which we've labeled as Equation 4, is a linear equation with two variables and represents a crucial intermediate result in our quest to solve the SPLTV system. It's simpler than the original equations and sets the stage for further steps in the solution process.

Step 3: Work with the Remaining Equations

Now we have two equations with 'x' and 'y':

• Equation 3: $3x - 2y = 12$
• Equation 4: $x - y = 5$

Let's solve for 'x' in Equation 4: $x = y + 5$

Substitute this into Equation 3:

$3(y + 5) - 2y = 12$

$3y + 15 - 2y = 12$

$y = -3$

When dealing with the remaining equations in the system, the focus shifts to manipulating these equations to solve for the remaining variables. In our scenario, after eliminating 'z', we were left with two equations involving 'x' and 'y'. These were Equation 3: $3x - 2y = 12$ and Equation 4: $x - y = 5$. The goal now is to use these two equations to find the values of 'x' and 'y'. A common strategy for solving such a system is substitution. This involves solving one equation for one variable and then substituting that expression into the other equation. In our case, we chose to solve Equation 4 for 'x'. This gave us $x = y + 5$. This expression tells us that 'x' is equal to 'y' plus 5. Now, we can substitute this expression for 'x' into Equation 3. This replaces 'x' in Equation 3 with 'y + 5', giving us $3(y + 5) - 2y = 12$. This equation now only involves the variable 'y', making it much easier to solve. We can simplify this equation by distributing the 3 and combining like terms: $3y + 15 - 2y = 12$. This further simplifies to $y + 15 = 12$. Subtracting 15 from both sides gives us $y = -3$. This means that the value of 'y' that satisfies the system of equations is -3. With the value of 'y' now known, we can use it to find the value of 'x'. We can substitute $y = -3$ back into the equation $x = y + 5$ to get $x = -3 + 5$, which simplifies to $x = 2$. So, the value of 'x' that satisfies the system of equations is 2. At this point, we have found the values of 'x' and 'y'. However, we still need to find the value of 'z'. To do this, we can substitute the values of 'x' and 'y' into one of the original equations that involves 'z'.

Step 4: Find the Value of 'x'

Substitute $y = -3$ into $x = y + 5$:

$x = -3 + 5$

$x = 2$

Finding the value of 'x' after determining the value of 'y' is a straightforward substitution process. In our case, we had already solved Equation 4 for 'x' in terms of 'y', giving us the equation $x = y + 5$. This equation provides a direct relationship between 'x' and 'y', allowing us to easily find 'x' once we know 'y'. In the previous step, we found that $y = -3$. This means that the value of 'y' that satisfies the system of equations is -3. Now, we can substitute this value of 'y' into the equation $x = y + 5$ to find the corresponding value of 'x'. Substituting $y = -3$ into $x = y + 5$ gives us $x = -3 + 5$. This simplifies to $x = 2$. Therefore, the value of 'x' that satisfies the system of equations is 2. At this point, we have found the values of 'x' and 'y'. However, we still need to find the value of 'z' to fully solve the system of equations. To do this, we can substitute the values of 'x' and 'y' into one of the original equations that involves 'z'. This will give us an equation with only 'z' as the unknown, which we can then solve to find the value of 'z'. It's important to choose an original equation that is relatively simple and easy to work with to minimize the chances of making errors in the substitution and simplification process. Once we have found the value of 'z', we will have found the values of all three variables, 'x', 'y', and 'z', and we will have successfully solved the system of equations.

Step 5: Solve for 'z'

Substitute $y = -3$ into the first original equation:

$5(-3) - 8z = -19$

$-15 - 8z = -19$

$-8z = -4$

$z = \frac{-4}{-8}$

$z = \frac{1}{2}$

To solve for 'z', we substitute the known value of 'y' into one of the original equations that contains 'z'. The first original equation, $5y - 8z = -19$, is a suitable choice for this purpose. By substituting $y = -3$ into this equation, we obtain $5(-3) - 8z = -19$. This simplifies to $-15 - 8z = -19$. The next step is to isolate the term containing 'z'. We can do this by adding 15 to both sides of the equation, which gives us $-8z = -19 + 15$. This further simplifies to $-8z = -4$. Finally, to solve for 'z', we divide both sides of the equation by -8, which gives us $z = \frac{-4}{-8}$. Simplifying this fraction, we find that $z = \frac{1}{2}$. Therefore, the value of 'z' that satisfies the system of equations is $\frac{1}{2}$. Now that we have found the values of 'x', 'y', and 'z', we have successfully solved the system of three-variable linear equations. The solution is $x = 2$, $y = -3$, and $z = \frac{1}{2}$. This solution represents the point in three-dimensional space where all three planes defined by the original equations intersect. It's important to check our solution by substituting these values back into the original equations to ensure that they all hold true. This helps to verify that we have not made any errors in our calculations and that our solution is correct. In this case, substituting $x = 2$, $y = -3$, and $z = \frac{1}{2}$ into the original equations confirms that they are all satisfied, so we can be confident that our solution is correct.

Step 6: The Solution

So, the solution to the system of equations is:

• $x = 2$
• $y = -3$
• $z = \frac{1}{2}$

There you have it! Solving SPLTV isn't so scary when you break it down into smaller steps. Practice makes perfect, so keep at it!

The solution to the system of equations consists of the values of the variables 'x', 'y', and 'z' that simultaneously satisfy all three equations. In our case, after performing the necessary steps of elimination, substitution, and simplification, we arrived at the solution $x = 2$, $y = -3$, and $z = \frac{1}{2}$. This means that when we substitute these values into the original equations, they will all hold true. To verify this, we can substitute these values back into the original equations: $5y - 8z = -19$, $5x - 8z = 6$, and $3x - 2y = 12$. For the first equation, $5y - 8z = -19$, substituting $y = -3$ and $z = \frac{1}{2}$ gives us $5(-3) - 8(\frac{1}{2}) = -15 - 4 = -19$, which is true. For the second equation, $5x - 8z = 6$, substituting $x = 2$ and $z = \frac{1}{2}$ gives us $5(2) - 8(\frac{1}{2}) = 10 - 4 = 6$, which is also true. For the third equation, $3x - 2y = 12$, substituting $x = 2$ and $y = -3$ gives us $3(2) - 2(-3) = 6 + 6 = 12$, which is also true. Since all three equations are satisfied by these values, we can confidently conclude that $x = 2$, $y = -3$, and $z = \frac{1}{2}$ is the correct solution to the system of equations. This solution represents the point in three-dimensional space where the three planes defined by the equations intersect. It's important to note that not all systems of equations have a unique solution. Some systems may have no solution (inconsistent systems), while others may have infinitely many solutions (dependent systems). However, in this case, we have found a unique solution that satisfies all three equations. This completes the process of solving the system of three-variable linear equations. Remember, practice is key to mastering this skill, so keep working on similar problems to improve your understanding and proficiency.