Question

Solve each system of equations using any method you wish. \(\left\\{\begin{array}{r}-4 x+3 y+2 z=6 \\ 3 x+y-z=-2 \\ x+9 y+z=6\end{array}\right.\)

Short answer

The solution is \(( x, y, z ) = (-1.5, 1, -1.5)\).

Step-by-step solution

Choose a Method

Decide which method to use to solve the system of equations. Options include substitution, elimination, or matrix methods. For this example, the elimination method will be used.

Write down the system of equations

\(\begin{align*}-4x + 3y + 2z &= 6 \ 3x + y - z &= -2 \ x + 9y + z &= 6 \end{align*}\)

Eliminate one variable

To start, eliminate the variable \(z\) by combining pairs of equations. First, add the second equation to the third equation:\( (3x + y - z) + (x + 9y + z) = -2 + 6 \)This simplifies to:\( 4x + 10y = 4 \)Next, multiply the second equation by 2 and add it to the first equation:\( 2(3x + y - z) = 2(-2) \)\( 6x + 2y - 2z = -4 \)\( -4x + 3y + 2z + 6x + 2y - 2z = 6 - 4 \)This simplifies to:\( 2x + 5y = 2 \)

Solve the simplified system

Now solve the simplified system of two equations:\(\begin{align*} 4x + 10y &= 4 \ 2x + 5y &= 2 \end{align*} \)Observe that the second equation is just half of the first equation. Therefore, solving one of these gives the relationships:\( 2x + 5y = 2 \)Divide through by 2:\( x + \frac{5}{2}y = 1 \)Solve for \(x\):\( x = 1 - \frac{5}{2}y \)

Substitute back to find specific values

Substitute \( x = 1 - \frac{5}{2}y \) back into the first or second simplified equation to find \(y\). For example, using the second simplified equation:\( 2(1 - \frac{5}{2}y) + 5y = 2 \)\( 2 - 5y + 5y = 2 \)This simplifies to: \(2 = 2\), which is always true, meaning \(y\) can be any real number. Substitute \(y = 1\) (arbitrary choice) back into \( x = 1 - \frac{5}{2}y \). This gives \( x = 1 - \frac{5}{2}(1) = -1.5 \). Now substitute \(x = -1.5\) and \(y = 1\) back into the original equations to solve for \(z\).

Find \(z\)

Using equation 2: \( 3(-1.5) + 1 - z = -2 \)\( -4.5 + 1 - z = -2 \)\( -3.5 - z = -2 \)\( z = -1.5 \)

Verify the solution

Substitute \(x = -1.5\), \(y = 1\), and \(z = -1.5\) back into all three original equations to ensure they hold true:1. \( -4(-1.5) + 3(1) + 2(-1.5) = 6 \) which simplifies to \( 6 = 6 \)2. \( 3(-1.5) + 1 - (-1.5) = -2 \) which simplifies to \( -2 = -2 \)3. \( -1.5 + 9(1) + (-1.5) = 6 \) which simplifies to \( 6 = 6 \)

Key concepts

Elimination Method

The elimination method is a powerful technique used to solve systems of linear equations by removing one variable at a time. This is achieved through strategically adding or subtracting equations. To start utilizing this method, follow these steps:

• Choose a variable to eliminate first. Look for the simplest coefficients to work with.
• Manipulate the equations so that when added or subtracted, one variable cancels out.
• Solve the resulting equation for the remaining variable(s).
• Repeat the steps until all variables are found.

In the given exercise, we removed the variable \(z\) by combining and manipulating the original equations. By carefully selecting and transforming the equations, we reduced the 3-variable system into a simpler 2-variable system. This subsequently made solving for specific values more manageable.

Linear Equations

A linear equation is an equation that forms a straight line when graphed. These are typically in the form \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants. Key characteristics of linear equations include:

• Each term is either a constant or the product of a constant and a single variable.
• No variables are raised to a power other than one.
• They graph as straight lines in coordinate space.

In our example, we have three linear equations:
\[-4x + 3y + 2z = 6\] \[3x + y - z = -2\] \[x + 9y + z = 6\]The goal is to find values of \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously. Through elimination and substitution methods, a consistent solution for these variables is determined.

Substitution Method

The substitution method provides an alternative approach to solving systems of equations by expressing one variable in terms of the others, then replacing it in the original equations. Here's how you can approach it:

• Solve one of the equations for one variable in terms of the other(s).
• Substitute this expression into the other equation(s).
• Solve the reduced equation for another variable.
• Substitute back to find the remaining variables.

This method simplifies complex systems by breaking them down step-by-step. For instance, in our exercise, after eliminating \(z\), we expressed \(x\) in terms of \(y\):
\[x = 1 - \frac{5}{2}y\]By substituting this expression back into the simplified equations, we determined specific values for \(x\) and \(y\), and finally solved for \(z\). While the process is systematic, it requires careful algebraic manipulation to ensure accuracy.