Problem 49 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{aligned} 4 x-3 y+z &=-8 \\ -2 x+y-3 z &=-4 \\ x-y+2 z &=3 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = -2 \), \( y = 1 \), \( z = 3 \).

Step 1: Define the System of Equations

The system of equations is given as follows: \[ \begin{aligned} 4x - 3y + z &= -8 \ -2x + y - 3z &= -4 \ x - y + 2z &= 3 \end{aligned} \] We need to solve for the variables \(x\), \(y\), and \(z\).

Step 2: Use Substitution or Elimination

First, let's focus on simplifying the equations. We can use elimination to eliminate variable \(x\) from the second equation. Multiply the third equation by 2 and add it to the second equation:\[2(x - y + 2z) = 2 \, ext{gives} \, 2x - 2y + 4z = 6\]Add this to the second equation:\[(-2x + y - 3z) + (2x - 2y + 4z) = -4 + 6\]Resulting in:\[-y + z = 2\]

Step 3: Solve for One Variable

Now solve \(-y + z = 2\) for \( z \) in terms of \( y \):\[ z = y + 2 \]

Step 4: Substitute Back and Solve

Substitute \(z = y + 2\) into the first and third equations to eliminate \(z\). First equation becomes:\[ 4x - 3y + (y + 2) = -8 \]Simplify to:\[ 4x - 2y = -10 \] Third equation becomes:\[ x - y + 2(y + 2) = 3 \, ext{or} \, x + y + 4 = 3 \]Simplify to:\[ x + y = -1 \]

Step 5: Solve for x and y

Now solve the two equations obtained:\[4x - 2y = -10 \ x + y = -1\] Solve the second equation for \(x\): \[ x = -1 - y \]Substitute into the first:\[ 4(-1 - y) - 2y = -10 \]Simplify and solve:\[ -4 - 4y - 2y = -10 \]\[ -6y = -6 \]\[ y = 1 \]

Step 6: Solve for Remaining Variables

Substitute \( y = 1 \) back into \( x = -1 - y \): \[ x = -1 - 1 = -2 \]Now, substitute \( y = 1 \) back into \( z = y + 2 \):\[ z = 1 + 2 = 3 \]

Step 7: Verify the Solution

Substitute \( x = -2 \), \( y = 1 \), and \( z = 3 \) back into the original equations to verify:First equation: \[ 4(-2) - 3(1) + 3 = -8 \, ext{True}\]Second equation: \[ -2(-2) + 1 - 3(3) = -4 \, ext{True}\]Third equation: \[ -2 - 1 + 2(3) = 3 \, ext{True}\] All equations are satisfied, confirming the solution is correct.

Key Concepts

Elimination Method

When solving linear systems, the elimination method is a powerful technique. Its main goal is to eliminate one variable from the equations, simplifying the system, so you can solve for the other variables more easily. To achieve this, you can add or subtract the equations after aligning them in a way that removes one of the variables. Often, you’ll need to multiply one or both equations by suitable numbers to make the coefficients of one variable opposite.

Let's illustrate this with a simple step from the given solution:

• Take the equation: \( -2x + y - 3z = -4 \)
• Multiply the equation \( x - y + 2z = 3 \) by 2 to make a manipulation possible: \( 2x - 2y + 4z = 6 \)
• Add both equations together to eliminate \( x \): \(-2x + y - 3z) + (2x - 2y + 4z) = -4 + 6 \)
• The result will be a simpler equation: \(-y + z = 2 \)

This process effectively narrows down the number of variables in an equation, allowing further simplification using other methods like substitution.

Substitution Method

In the substitution method, you solve one of the equations for one variable, then substitute that expression in the other equations. This way, you gradually reduce the system down to simpler equations with fewer variables. Let's break this down using the solution process highlighted in the example:

Start by solving the simplified equation \( -y + z = 2 \) for \( z \):

• This gives \( z = y + 2 \)
• Substitute \( z = y + 2 \) into the first equation \( 4x - 3y + z = -8 \), yielding \( 4x - 3y + (y+2) = -8 \)
• Simplify this to \( 4x - 2y = -10 \)

The substitution simplifies complex systems by keeping track of each variable’s value through its relationship with another. Ultimately, substitution lets you find one variable at a time by reducing the number of unknowns required to solve each equation, making it a friendlier approach to tackling multivariable problems.

Verify Solution

After solving a system of linear equations, it is essential to verify that your solutions are correct. This step ensures that your answers satisfy all the original equations. Let’s discuss how you can effectively verify your solutions using the example provided.

Take the final computed values \( x = -2 \, \ y = 1 \, \ z = 3 \), and substitute these into each original equation to verify:

• First equation: \( 4(-2) - 3(1) + 3 = -8 \)
• Second equation: \( -2(-2) + 1 - 3(3) = -4 \)
• Third equation: \( -2 - 1 + 2(3) = 3 \)

Each equation, when tested, is satisfied or becomes true. Therefore, the original set of equations is completely satisfied, affirming your solution as correct. Verification not only boosts confidence in your answer but also helps spot potential missteps, ensuring accuracy in problem-solving.