Solving 3x3 Systems Of Equations Elimination And Substitution Methods

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In mathematics, solving systems of linear equations is a fundamental skill with applications across various fields, including engineering, physics, economics, and computer science. A system of linear equations consists of two or more equations with the same set of variables. The solution to a system of equations is a set of values for the variables that satisfy all equations simultaneously. In this comprehensive guide, we will delve into the methods for solving a $3 \times 3$ system of linear equations, providing a step-by-step approach and illustrative examples.

Understanding $3 \times 3$ Systems of Equations

A $3 \times 3$ system of linear equations involves three equations with three unknown variables, typically denoted as $x$, $y$, and $z$. Each equation represents a plane in three-dimensional space, and the solution to the system corresponds to the point where all three planes intersect. A $3 \times 3$ system can have one unique solution, infinitely many solutions, or no solution, depending on the relationships between the planes.

Let's consider the following system of equations as an example:

$\begin{aligned} x + 2y - z &= -3 \\ 2x - y + z &= 5 \\ x - y + z &= 4 \end{aligned}$

Our goal is to find the values of $x$, $y$, and $z$ that satisfy all three equations simultaneously. We will explore two primary methods for solving such systems: the elimination method and the substitution method.

Method 1: Elimination Method

The elimination method involves systematically eliminating variables from the equations until we are left with a single equation with one unknown variable. This is achieved by adding or subtracting multiples of the equations to cancel out the desired variables. Let's apply the elimination method to our example system:

Step 1: Eliminate $x$ from Equations 2 and 3

To eliminate $x$ from Equation 2, we can multiply Equation 1 by -2 and add it to Equation 2:

$\begin{aligned} -2(x + 2y - z) &= -2(-3) \\ -2x - 4y + 2z &= 6 \end{aligned}$

Adding this modified equation to Equation 2:

$\begin{aligned} (-2x - 4y + 2z) + (2x - y + z) &= 6 + 5 \\ -5y + 3z &= 11 \end{aligned}$

Now, let's eliminate $x$ from Equation 3. Multiply Equation 1 by -1 and add it to Equation 3:

$\begin{aligned} -1(x + 2y - z) &= -1(-3) \\ -x - 2y + z &= 3 \end{aligned}$

Adding this modified equation to Equation 3:

$\begin{aligned} (-x - 2y + z) + (x - y + z) &= 3 + 4 \\ -3y + 2z &= 7 \end{aligned}$

We now have a reduced system of two equations with two variables:

$\begin{aligned} -5y + 3z &= 11 \\ -3y + 2z &= 7 \end{aligned}$

Step 2: Eliminate $y$ from the Reduced System

To eliminate $y$, multiply the first equation by -3 and the second equation by -5:

$\begin{aligned} -3(-5y + 3z) &= -3(11) \\ 15y - 9z &= -33 \end{aligned}$

$\begin{aligned} -5(-3y + 2z) &= -5(7) \\ 15y - 10z &= -35 \end{aligned}$

Subtract the second equation from the first:

$\begin{aligned} (15y - 9z) - (15y - 10z) &= -33 - (-35) \\ z &= 2 \end{aligned}$

Step 3: Back-Substitute to Find $y$ and $x$

Now that we have found $z = 2$, we can substitute it back into one of the equations in the reduced system to find $y$. Let's use the equation $-3y + 2z = 7$:

$\begin{aligned} -3y + 2(2) &= 7 \\ -3y + 4 &= 7 \\ -3y &= 3 \\ y &= -1 \end{aligned}$

Finally, substitute $y = -1$ and $z = 2$ into Equation 1 to find $x$:

$\begin{aligned} x + 2(-1) - 2 &= -3 \\ x - 2 - 2 &= -3 \\ x - 4 &= -3 \\ x &= 1 \end{aligned}$

Therefore, the solution to the system of equations is $x = 1$, $y = -1$, and $z = 2$.

Method 2: Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equations. This process is repeated until we have a single equation with one unknown variable. Let's apply the substitution method to our example system:

Step 1: Solve Equation 1 for $x$

$\begin{aligned} x + 2y - z &= -3 \\ x &= -3 - 2y + z \end{aligned}$

Step 2: Substitute into Equations 2 and 3

Substitute the expression for $x$ into Equation 2:

$\begin{aligned} 2(-3 - 2y + z) - y + z &= 5 \\ -6 - 4y + 2z - y + z &= 5 \\ -5y + 3z &= 11 \end{aligned}$

Substitute the expression for $x$ into Equation 3:

$\begin{aligned} (-3 - 2y + z) - y + z &= 4 \\ -3 - 2y + z - y + z &= 4 \\ -3y + 2z &= 7 \end{aligned}$

We again obtain the reduced system of two equations with two variables:

$\begin{aligned} -5y + 3z &= 11 \\ -3y + 2z &= 7 \end{aligned}$

Step 3: Solve the Reduced System

We can solve this reduced system using either elimination or substitution. Let's use elimination again. Multiply the first equation by -3 and the second equation by -5:

$\begin{aligned} -3(-5y + 3z) &= -3(11) \\ 15y - 9z &= -33 \end{aligned}$

$\begin{aligned} -5(-3y + 2z) &= -5(7) \\ 15y - 10z &= -35 \end{aligned}$

Subtract the second equation from the first:

$\begin{aligned} (15y - 9z) - (15y - 10z) &= -33 - (-35) \\ z &= 2 \end{aligned}$

Step 4: Back-Substitute to Find $y$ and $x$

Substitute $z = 2$ into the equation $-3y + 2z = 7$:

$\begin{aligned} -3y + 2(2) &= 7 \\ -3y + 4 &= 7 \\ -3y &= 3 \\ y &= -1 \end{aligned}$

Substitute $y = -1$ and $z = 2$ into the expression for $x$:

$\begin{aligned} x &= -3 - 2(-1) + 2 \\ x &= -3 + 2 + 2 \\ x &= 1 \end{aligned}$

Therefore, the solution to the system of equations is $x = 1$, $y = -1$, and $z = 2$, which matches the solution obtained using the elimination method.

Practice Problem

Solve the following system of equations using either the elimination or substitution method:

$\begin{aligned} 2x + y - z &= 1 \\ x - y + 2z &= 5 \\ 3x + 2y + z &= 0 \end{aligned}$

Conclusion

Solving $3 \times 3$ systems of linear equations is a crucial skill in mathematics with broad applications. The elimination and substitution methods provide systematic approaches to finding the solutions. By mastering these methods, you can confidently tackle various problems involving systems of linear equations in diverse fields.

Keywords: system of equations, linear equations, elimination method, substitution method, $3 \times 3$ system, solving equations, mathematics